Notes&Solutions.txt

Equality Laws
		Substitution
E
=====
E[v:=F]

		Reflexivity
x=x

		Symmetry
(x=y)=(y=x)

		Transitivity
X=Y,Y=Z
======
X=Z

		Leibniz
X=Y
============
E[z:=X]=E[z:=Y]

Propositional Calculus Inference Rules
		Substitution
P
=====
P[r:=Q]

		Transitivity
P=Q,Q=R
=======
P=R

		Leibniz
P=Q
============
E[r:=P]=E[r:=Q]

ARSENAL:
1.1 Substitution: LOOKUP
1.2 Reflexivity: LOOKUP
1.3 Symmetry: LOOKUP
1.4 Transitivity: LOOKUP
1.5 Leibniz: LOOKUP
1.6 g(z) = 3•z+6
1.7 g.z: E
1.8 Leibniz: \frac{X=Y}{g.X=g.Y}
1.9 2•x/2=x
1.10 x:=E
1.11 {P}S{Q}
1.12 Definition of assignment: {R[x:=E]}x:=E{R}
2.1 Satisfiability,Validity Definition. A boolean expression P is satisfied in 
a state 
	  if its value is true in that state; P is satisfiable if there is a state 
in which it is satisfied;
	  and P is valid if it is satisfied in every state. A valid boolean 
expression is called a tautology.
2.2 Duality Definition. The dual PD of a boolean expression P is constructed 
from P by interchanging
	  occurrences of
			1. true and false,
			2. ∧ and ∨,
			3. ≡ and ≢,
			4. → and ⇍, and
			5. ⇐ and ⇏
2.3 Metatheorem Duality
	  1. P is valid iff ¬PD is valid.
	  2. P ≡ Q is valid iff PD ≡ QD is valid.
2.4 Henry VIII had one son and Cleopatra had two.
	  p: Henry VIII had one son and Cleopatra had two
		  A boolean variable that can denote a proposition is sometimes called 
a  propositional variable,
		  but we will stick to the term boolean variable.
			  x: Henry VIII had one son,
			  y: Cleopatra had two (sons),
		  English statement: x and y
		  Boolean expression: x ∧ y
2.5 Translation into a boolean expression.
	  To translate proposition p into a boolean expression:
		  1. Introduce boolean variable to denote subpropositions.
		  2. Replace these subpropositions by their corresponding boolean 
variables.
		  3. Translate the result of step 2 into a boolean expression, using 
“obvious” translation 
			  of the English wordsinto oeprators.
			  | it is not the case that | ¬ |
			  | if p then q | p→q |
			  | and, but | ∧ |
			  | or | ∨ |
			  | not | ¬ |
2.4 Follow up
	  x: Henry VIII had one son,
	  y: Cleopatra had two (sons),
	  z: I’ll eat my hat,
	  w: 1 is prime.
		  | proposition | translation |
		  | Henry VIII had one son or I’ll eat my hat. | x∨z |
		  | Henry VIII had one son and 1 is not prime. | x∧¬w |
		  | If 1 is prime and Cleopatra had two sons, I’ll eat my hat. | w∧y→z |
3.1 Axiom, Associativity of ≡: ((p ≡ q) ≡ r) ≡ ( p ≡ (q ≡ r))
3.2 Axiom, Symmetry of ≡: p ≡ q ≡ q ≡ p
3.3 Axiom, Identity of≡: true ≡ q ≡ q
3.4 True
	=    <Identity of ≡(3.3), with q:=true>
		true ≡ true
	=    <Identity of ≡(3.3) —replace the second true>
		true ≡ q ≡ q
3.5 Reflexivity of ≡: p ≡ p
	p ≡ p
	=    <Symmetry of ≡(3.2), with q:=p —replace p≡p by p≡q≡q≡p>
		p ≡ q ≡ q ≡ p —Symmetry of ≡(3.2)
3.6 Proof method. To prove that P ≡ Q is a theorem, transform P to Q or Q to P 
using Leibniz.
3.7 Metatheorem. Any two theorems are equivalent.
3.8 Axiom, Definition of false: false ≡ ¬true
3.9 Axiom, Distributivity of ¬ over ≡ : ¬(p ≡ q) ≡ ¬p ≡ q
3.10 Axiom, Definition of ≢ : (p ≢ q) ≡ ¬(p ≡ q)
3.11 ¬p ≡ q ≡ p ≡ ¬q
	 =    <(3.9), ¬(p ≡ q) ≡ ¬p ≡ q>
		¬(p ≡ q) ≡ p ≡ ¬q
	 =    <(3.9), with p,q:=q,p    —i.e. ¬(q ≡ p) ≡ ¬q ≡ p>
		¬(p ≡ q) ≡ ¬(p ≡ q)    —Reflexivity of ≡ (3.5)
3.12 Double negation: ¬¬p ≡ p
3.13 Negation of false: ¬false ≡ true
3.14 (p ≢ q) ≡ ¬p ≡ q
3.15 ¬p ≡ p ≡ false
3.16 Symmetry of ≢: (p ≢ q) ≡ (q ≢ p)
	p ≢ q
	=    <Def. of ≢ (3.10)>
		¬(p ≡ q)
	=    <Symmetry of ≡ (3.2)>
		¬(q ≡ p)
	=    <Def. of ≢ (3.10), with p,q:=q,p>
		q ≢ p
3.17 Associativity of ≢: ((p ≢ q) ≢ r) ≡ (p ≢ (q ≢ r))
3.18 Mutual associativity: ((p ≢ q) ≡ r) ≡ (p ≢ (q ≡ r))
3.19 Mutual interchangeability: p ≢ q ≡ r ≡ p ≡ q ≢ r
3.20 Heuristic. Identify applicable theorems by matching the structure of 
expressions or subexpressions.
		The operators that appear in a boolean expression and the shape of its 
subexpressions can focus the
		choice of theorems to be used in manipulating it.
3.21 P0 ≡ P1 ≡ … ≡ Pn is true exactly when an even number of the Pi are false.
3.22 Principle: Structure proofs to avoid repeating the same subexpression on 
many lines.
3.23 Heuristic of Definition Elimination: To prove a theorem concerning an 
operator o that is defined in 
		terms of another, say •, expand the definition of o to arrive at a 
formula that contains •;
		exploit properties of • to manipulate the formula; and then (possibly) 
reintroduce o using its definition.
3.24 Axiom, Symmetry of ∨: p ∨ q ≡ q ∨ p
3.25 Axiom, Associativity of ∨: (p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
3.26 Axiom, Idempotency of ∨: p ∨ p ≡ p
3.27 Axiom, Distributivity of ∨ over ≡: p ∨ (q ≡ r)  ≡ p ∨ q ≡ p ∨ r
		p ∨ (q ≡ r)  ≡ p ∨ q ≡ p ∨ r
		Replacing LHS by the RHS called “multiplying out”.
		Replacing RHS by the LHS called “factoring”.
3.28 Axiom, Excluded middle: p ∨ ¬p
		Axiom Excluded middle can be interpreted to mean that in any state 
either p or ¬p is true;
		there is no middle ground.
3.29 Zero of ∨: p ∨ true ≡ true
	 Z is a zero of a binary operation o if x o Z = Z o x = Z, for all x.
	 Z is a left zero if Z o x = Z, for all x. Z is a right zero if x o Z = Z, for all x.
	 The term zero comes from the fact that 0 is the zero of •.
	 p ∨ true
	 =    <Identity of ≡(3.3), with q:=p —replace true by p≡p>
		p ∨ (p ≡ p)
	 =    <Distributivity of ∨ over ≡(3.27), with q,r:=p,p —replace p∨(p≡p) by p∨p≡p∨p>
		p ∨ p ≡ p ∨ p
	 =    <Identity of ≡(3.3), with q:=p∨p —replace p∨p≡p∨p by true>
		true
3.30 Identity of ∨: p ∨ false ≡ p
	true
	=    <Identity of ≡(3.3), with q:=true —replace true by true≡true>
		true ≡ true
	=    <Identity of ≡(3.3) with q:=p —replace true by p≡p>
		true ≡ p ≡ p
	=    <Idempotency of ∨(3.26) —replace p by p∨p>
		true ≡ p ∨ p ≡ p
	=    <Excluded middle(3.28) —replace true by p∨¬p>
		p ∨ ¬p ≡ p ∨ p ≡ p
	=    <Distributivity of ∨ over ≡(3.27), with q:=p
		  —replace p∨¬p≡p∨p≡p by p∨(¬p≡p)>
		p ∨ (¬p ≡ p) ≡ p
	=    <Distributivity of ¬ over ≡(3.9), with q:=p —replace (¬p≡p) by ¬(p≡p)>
		p ∨ ¬(p ≡ p) ≡ p
	=    <Reflexivity of ≡(3.5) —replace p≡p by true>
		p ∨ ¬true ≡ p
	=    <Def. of false(3.8) —replace ¬true by false>
		p ∨ false ≡ p
3.31 Distributivity of ∨ over ∨: p ∨ (q ∨ r) ≡ (p ∨ q) ∨ (p ∨ r)
3.32 p ∨ q ≡ p ∨ ¬q ≡ p
	p ∨ q ≡ p ∨ ¬q
	=    <Distributivity of ∨ over ≡(3.27), with r:=¬q>
		p ∨ (q ≡ ¬q)
	=    <Symmetry of ≡(3.2), with p,q:=q,¬q>
		p ∨ (¬q ≡ q)
	=    <Distributivity of ¬ over ≡(3.9), with p:=q>
		p ∨ ¬(q ≡ q)
	=    <Identity of ≡(3.3)>
		p ∨ ¬(true)
	=    <Def. of false(3.8)>
		p ∨ false —theorem(3.30)
3.33 Heuristic: To prove P ≡ Q, transform the expression with the most 
structure (either P or Q) into the other.
3.34 Principle: Structure proofs to minimize the number of rabbits pulled out 
of a hat —make each
		step obvious, based on the structure of the expression and the goal of 
the manipulation.
3.35 Axiom, Golden rule: p ∧ q ≡ p ≡ q ≡ p ∨ q
		It can also be rewritten as (p ≡ q) ≡ (p ∧ q  ≡ p ∨ q)
3.36 Symmetry of ∧: p ∧ q ≡ q ∧ p
3.37 Associativity of ∧: (p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
	(p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
		(p ∧ q) ∧ r
		=    <Golden rule(3.35), —replace p∧q with (p≡q≡p∨q)>
			(p ≡ q ≡ p ∨ q) ∧ r
		=    <Golden rule(3.35) with p,q:=p≡q≡p∨q,r —replace (p≡q≡p∨q)∧r by
			  p≡q≡p∨q≡r≡(p≡q≡p∨q)∨r>
			p ≡ q ≡ p ∨ q ≡ r ≡ (p ≡ q ≡ p ∨ q) ∨ r
		=    <Symmetry of ∨(3.24), with p,q:=(p≡q≡p∨q),r —replace (p≡q≡p∨q)∨r
			  by r∨(p≡q≡p∨q)>
			p ≡ q ≡ p ∨ q ≡ r ≡ r ∨ (p ≡ q ≡ p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p≡q,p∨q —replace
			  r∨(p≡q≡p∨q) by r∨(p≡q)≡r∨(p∨q)>
			p ≡ q ≡ p ∨ q ≡ r ≡ r ∨ (p ≡ q) ≡ r ∨ (p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p,q —replace
			  r∨(p≡q) by r ∨ p ≡ r ∨ q>
			p ≡ q ≡ p ∨ q ≡ r ≡ r ∨ p ≡ r ∨ q ≡ r ∨ (p ∨ q)
		=    <Distributivity of ∨ over ∨(3.31), with p,q,r:=r,p,q —replace
			  r∨(p∨q) by r∨p∨r∨q)
			p ≡ q ≡ p ∨ q ≡ r ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ r ∨ q
		=    <Symmetry and associativity of ≡ and ∨>
			p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p ≡ p ∨ q ∨ r
			p ≡ q ≡ q ∨ r ≡ r ∨ p ≡ (p ∨ q) ∧ r
		p ∧ (q ∧ r)
		=    <Symmetry of ∧ (3.36)>
			(q ∧ r) ∧ p
		=    <(3.55), with p,q,r:=q,r,p>
			q ≡ r ≡ p ≡ q ∨ r ≡ r ∨ p ≡ p ∨ q ≡ q ∨ r ∨ v
		=    <Symmetry and associativity of ≡ and ∨>
			p ≡ q ≡ p ∨ q ≡ r ≡ p ∨ r≡ q ∨ r ≡ p ∨ q ∨ r
		=    <(3.55)>
			(p ∧ q) ∧ r
3.38 Idempotency of ∧: p ∧ p ≡ p
3.39 Identity of ∧: p ∧ true ≡ p
	 p ∧ true ≡ p
	 =    <Golden rule(3.35) with q:=true —replace two equivalents>
		p ∨ true ≡ true
3.40 Zero of ∧: p ∧ false ≡ false
3.41 Distributivity of ∧ over ∧: p ∧ (q ∧ r) ≡ (p ∧ q) ∧ (p ∧ r)
3.42 Contradiction: p ∧ ¬p ≡ false
3.43 Absorption:
		1. p ∧ (p ∨ q) ≡ p
		2. p ∨ (p ∧ q) ≡ p
3.44 Absorption:
		1. p ∧ (¬p ∨ q) ≡ p ∧ q
		   p ∧ (¬p ∨ q)
		   =    <Golden rule(3.35), with q:=¬p∨q>
			   p ≡ ¬p ∨ q ≡ p ∧ ¬p ∨ q
		   =    <Excluded middle(3.28)>
			   p ≡ ¬p ∨ q ≡ true ∨ q
		   =    <(3.29), true ∨ p ≡ true>
		2. p ∨ (¬p ∧ q) ≡ p ∨ q
3.45 Distributivity of ∨ over ∧:p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
3.46 Distributivity of ∧ over ∨: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
3.47 De Morgan:
		1. ¬(p ∧ q) ≡ ¬p ∨ ¬q
		2. ¬(p ∨ q) ≡ ¬p ∧ ¬q
3.48 p ∧ q ≡ p ∧ ¬q ≡ ¬p
3.49 p ∧ (q ≡ r) ≡ p ∧ q ≡ p ∧ r ≡ p
	 p ∧ (q ≡ r) ≡ p
	 =    <Golden rule, with q:=q≡r —replace two equivalents>
		p ∨ (q ≡ r) ≡ q ≡ r
	 =    <Distributivity of ∨ over ≡(3.27)>
		p ∨ q ≡ p ∨ r ≡ q ≡ r
	 =    <Symmetry of ≡(3.2)>
		p ∨ q ≡ q ≡ p ∨ r ≡ r
	 =    <Golden rule, twice —replace p∨q≡q and p∨r≡r>
		p ∧ q ≡ p ≡ p ∧ r ≡ p
	 =    <Symmetry of ≡(3.2), with q:=p∨q≡p∨r>
		p ∧ q ≡ p ∧ r
3.50 p ∧ (q ≡ p) ≡ p ∧ q
3.51 Replacement: (p ≡ q) ∧ (r ≡ p) ≡ (p ≡ q) ∧ (r ≡ q)
3.52 Definition of ≡: p ≡ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)
3.53 Exclusive or: p ≢ q ≡ (¬p ∧ q) ∨ (p ∧ ¬q)
3.54 Principle: Lemmas can provide structure, bring to light interesting facts, and ultimately shorten a proof.
3.55 (p ∧ q) ∧ r ≡ p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p ≡ p ∨ q ∨ r
3.56 Heuristic: Exploit the ability to parse theorems like the Golden rule in many different ways.
3.57 Axiom, Definition of Implication:
	 p → q ≡ p ∨ q ≡ q
3.58 Axiom, Consequence:
	 p ⇐ q ≡ q → p
3.59 Definition of Implication: p → q ≡ ¬p ∨ q
3.60 Definition of Implication: p → q ≡ p ∧ q ≡ p
3.61 Contrapositive: p → q ≡ ¬q → ¬p
3.62 p → (q ≡ r) ≡ p ∧ q ≡ p ∧ r
	 p → (q ≡ r)
	 =   <Definition of implication(3.60)>
		p ∧ (q ≡ r) ≡ p
	 =   <(3.49), p ∧ (q ≡ r) ≡ p ∧ q ≡ p ∧ r ≡ p>
		p ∧ q ≡ p ∧ r
3.63 Distributivity of → over ≡: p → (q ≡ r) ≡ p → q ≡ p → r
3.64 p → (q → r) ≡ (p → q) → (p → r)
3.65 Currying: p ∧ q → r ≡ p → (q → r)
3.66 p ∧ (p → q) ≡ p ∧ q
3.67 p ∧ (q → p) ≡ p
3.68 p ∨ (p → q) ≡ true
3.69 p ∨ (q → p) ≡ q → p
3.70 p ∨ q → p ∧ q ≡ p ≡ q
3.71 Reflexivity of →: p → p ≡ true
3.72 Right zero of →: p → true ≡ true
3.73 Left identity of →: true → p ≡ p
3.74 p → false ≡ ¬p
3.75 false → p ≡ true
3.76 Weakening/strengthening:
	(a) p → p ∨ q Weakening
	(b) p ∧ q → p Weakening
	(c) p ∧ q → p ∨ q Weakening
	(d) p ∨ (q ∧ r) → p ∨ q Weakening
	(e) p ∧ q → p ∧ (q ∨ r) Weakening
The more rows in the truth table that are labeled as "false," the stronger it becomes.
3.77 Modus ponens: p ∧ (p → q) → q
3.78 (p → r) ∧ (q → r) ≡ (p ∨ q → r)
3.79 (p → r) ∧ (¬p → r) ≡ r
3.80 Mutual implication: (p → q) ∧ (q → p) ≡ p ≡ q
	(p → q) ∧ (q → p)
	=    <Definition of implication(3.59), twice>
		(¬p ∨ q) ∧ (¬q ∨ p)
	=    <Distributivity of ∧ over ∨(3.46), twice>
		(¬p ∧ ¬q) ∨ (¬p ∧ p) ∨ (q ∧ ¬q) ∨ (q ∧ p)
	=    <Contradiction(3.42), twice; Identity of ∨(3.30), twice>
		(¬p ∧ ¬q) ∨ (q ∧ p)
	=    <Alternative definition of ≡(3.52)>
		p ≡ q
3.81 Antisymmetry: (p → q) ∧ (q → p) → (p ≡ q)
3.82 Transitivity:
	(a) (p → q) ∧ (q → r) → (p → r)
	=    <Currying(3.65), with p,q:=(p → q) ∧ (q → r),p —to shunt the p in the consequent to the antecedent>
	p ∧ (p → q) ∧ (q → r) → r
	=    <(3.66) —replace first two conjuncts>
	p ∧ q ∧ (q → r) → r
	=    <3.66 —again, to replace second and third conjuncts —with p,q:=q,r>
	p ∧ q ∧ r → r —Strengthening(3.76b)
	(b) (p ≡ q) ∧ (q → r) → (p → r)
	(c) (p → q) ∧ (q ≡ r) → (p → r)
3.83 Axiom, Leibniz: (e = f) → (E[z:=e]=E[z:=f]) (E any expression)
	Inference rule Leibniz: "if X = Y is valid, i.e. true in all states, then so is E[z:=X] = E[z:=Y]".
	Axiom(3.83): "if e = f is true in a state, then E[z:=e] = E[z:=f] is true in that state.".
		diff: E=false∧z,e=true,f=false
			E[z:=e]=E[z:=f] is true but e = f is false.
3.84 Substitution:
	(a) (e = f) ∧ E[z:=e] ≡ (e = f) ∧ E[z:=f]
	(b) (e = f) → E[z:=e] ≡ (e = f) → E[z:=f]
	(c) q ∧ (e = f) → E[z:=e] ≡ q ∧ (e = f) → E[z:=f]
3.85 Replace by true:
	(a) p → E[z:=p] ≡ p → E[z:=true]
	(b) q ∧ p → E[z:=p] ≡ q ∧ p → E[z:=true]
3.86 Replace by false:
	(a) E[z:=p] → p ≡ E[z:=false] → p
	(b) E[z:=p] → p ∨ q ≡ E[z:=false] → p ∨ q
3.87 Replace by true: p ∧ E[z:=p] ≡ p ∧ E[z:=true]
3.88 Replace by false: p ∨ E[z:=p] ≡ p ∨ E[z:=false]
3.89 Shannon: E[z:=p] ≡ (p ∧ E[z:=true]) ∨ (¬p ∧ E[z:=false])
3.1 We have defined ≡ using three axioms. Assuming that the symbol true is identified
	with the symbol true of the previous chapter on boolean expressions, do the axioms
	uniquely identify operator ≡? Answer the question by seeing which of the 16 possible
	binary operators o (say) given in the truth table on page 26 satisfy ((p o q) o r) o (p o (q o r)),
	p o q q o p, and true o q q. (For example, the operator given by the last column does not
	satisfy true o q o q, since the operator always yields f.)
		Answer: The axioms indeed uniquely determine the operator ≡.
3.2 Use truth tables to show that axiom (3.1), (3.2), and (3.3) are valid (true in every state).
	Answer:
		|p|q|r|3-1|3-2|3-3|
		|t|t|t| t | t | t |
		|f|f|f| t | t | t |
		|t|f|f| t | t | t |
		|f|t|t| t | t | t |
		|f|t|f| t | t | t |
		|t|f|t| t | t | t |
		|f|f|t| t | t | t |
		|t|t|f| t | t | t |
		Based on the truth table analysis, it is confirmed that
		axioms (3.1), (3.2), and (3.3) hold true in every state,
		as their corresponding logical expressions consistently
		evaluate to true across all possible combinations of truth values.
3.3 Prove Reflexivity of ≡(3.5), p ≡ p.
	Answer:
		p ≡ p
		=    <Symmetry of  ≡(3.2) —replace first p by p≡q≡q>
				P:p
				Q:p≡q≡q
				E:r≡p
			p ≡ q ≡ q ≡ p —Symmetry of ≡(3.2)
3.4 Prove the following metatheorm. Q ≡ true is a theorem iff Q is a theorem.
	Answer:
		|Q|true|Q ≡ true|
		|t|   t|       t|
		|f|   t|       f|
		By examining the truth table for the given metatheorem, we can
		conclude that the statement "Q ≡ true is a theorem iff Q is a
		theorem" is indeed valid.
3.5 Prove the following metatheorem. Any two theorems are equivalent.
	Answer:
		true ≡ true
		=    <Identity of ≡(3.3) —replace second true by q=q>
				P:true
				Q:q≡q
				E:true≡r
			true ≡ q ≡ q —Identity of ≡(3.3)
3.6 Assume that operator ≡ is identified with operator ≡ of Sec. 2.1
	(see Exercise 3.1) and true is identified with the symbol true of Sec. 2.1.
	Prove that axioms (3.8) and (3.9) uniquely define operator ¬.
	That is, determine which of the four prefix operators o defined in the
	truth table on page 26 satisfy false ≡ otrue and o(p ≡ q) ≡ op ≡ q.
	Answer:
		|0|1|2|3|4|false≡¬true|¬(p≡q)≡¬p≡q|
		|t|t|t|f|f|     t     |     t     |
		|f|t|f|t|f|     t     |     t     |
		Through an analysis of the truth table for the four prefix operators,
		it becomes evident that column number 1 distinctly determines the operator ¬.
3.7 Prove theorem (3.11) in three different ways: start with ¬p ≡ q and transform
	it to p ≡ ¬q, start with ¬p ≡ p and transform it into q ≡ ¬q, and start with
	¬p and transform it into q ≡ p ≡ ¬q. Compare these three proofs and the one
	given on page 47. Which is simpler or shorter?
	Answer: The first one is simpler and shorter as shown below.
		¬p ≡ q
	=   <Distributivity of ¬ over ≡(3.9)>
		¬(q ≡ p)
	=   <Distributivity of ¬ over ≡(3.9)>
		p ≡ ¬q

		¬p ≡ p
	=    <Distributivity of ¬ over ≡(3.9)>
		¬(p ≡ p)
	=    <Reflexivity of ≡(3.5)>
		¬true
	=    <Identity of ≡(3.3)>
		¬(q ≡ q)
	=    <Distributivity of ¬ over ≡(3.9)>
		q ≡ ¬q

		¬p
	=    <Symmetry of ≡(3.2) —replace p by p≡q≡q>
		¬(p≡q≡q)
	=    <Distributivity of ¬ over ≡(3.9)>
		q ≡ p ≡ ¬q
3.8 Prove double negation (3.12), ¬¬p≡p
	Answer:
		¬¬p ≡ p
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=¬p,p —replace ¬¬p≡p
			  by ¬(¬p≡p)>
			¬(¬p ≡ p)
		=    <Symmetry of ≡(3.2), with p,q:=¬p,p —replace ¬p≡p by p≡¬p>
			¬(p ≡ ¬p)
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=p,¬p —replace ¬(p≡¬p)
			  by ¬p≡¬p>
			¬p ≡ ¬p
		=    <Reflexivity of ≡(3.5), with p:=¬p —replace ¬p≡¬p by p≡p>
			p ≡ p
3.9 Prove Negation of false (3.13), ¬false ≡ true
	Answer:
		¬false ≡ true
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=false,true>
			¬(false ≡ true)
		=    <Symmetry of ≡(3.2), with p,q:=false,true>
			¬(true ≡ false)
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=true,false>
			¬true ≡ false
		=    <Def. of false(3.8) —replace ¬true by false>
			false ≡ false
		=    <Metatheorem(3.7), Reflexivity of ≡(3.5), with p:= false and 
			  true(3.4) —replace false≡false by true
			true
3.10 Prove theorem (3.14), (p ≢ q) ≡ ¬p ≡ q
	 Answer:
		(p ≢ q)
		=    <Def. of ≢(3.10)>
			¬(p ≡ q)
		=    <Distributivity of ¬ over ≡(3.9)>
			¬p ≡ q
3.11 Prove theorem (3.15) by transforming ¬p ≡ p ≡ false to true using (3.11).
	 The proof should require only two uses of Leibniz.
	 Answer:
	 ¬p ≡ p ≡ false
	 =   <Def. of false(3.8) —replace false by ¬true>
		¬p ≡ p ≡ ¬true
	 =   <Identity of true(3.3) —replace true by q≡q>
		¬p ≡ p ≡ ¬(q ≡ q)
	 =   <Distributivity of ¬ over ≡(3.9)>
		¬p ≡ p ≡ ¬q ≡ q
	 =   <Symmetry of ≡(3.2), with p,q:=p≡¬q,q —replace p≡¬q≡q by q≡p≡¬q>
		¬p ≡ q ≡ p ≡ ¬q —theorem(3.11)

3.12 Prove Associativity of ≢(3.17), ((p ≢ q) ≢ r) ≡ (p ≢ (q ≢ r)), using the heuristic
	 of Definition elimination (3.23) —by eliminating ≢, using a property of ≡,
	 and introducing ≢.
	 Answer:
		((p ≢ q) ≢ r) ≡ (p ≢ (q ≢ r))
		=    <Def. of ≢(3.10) —replace (p≢q) by ¬(p≡q)>
			(¬(p ≡ q)≢ r) ≡ (p ≢ (q ≢ r))
		=    <Def. of ≢(3.10) with p,q:=q,r —replace (q≢r) by ¬(q≡r)>
			(¬(p ≡ q)≢ r) ≡ (p ≢ ¬(q ≡ r))
		=    <Def. of ≢(3.10) with p,q:¬(p≡q),r —replace ¬(p≡q)≢r by ¬(¬(p≡q)≡r)>
			¬(¬(p ≡ q) ≡ r) ≡ (p ≢ ¬(q ≡ r))
		=    <Def. of ≢(3.10) with q:¬(q≡r) —replace p≢¬(q≡r) by ¬(p≡¬(q≡r))>
			¬(¬(p ≡ q) ≡ r) ≡ ¬(p ≡ ¬(q ≡ r))
		=    <Symmetry of ≡(3.2), with p,q:=¬(¬(p≡q)≡r),¬(p≡¬(q≡r))>
			¬(p ≡ ¬(q ≡ r)) ≡ ¬(¬(p ≡ q) ≡ r)
		=    <Def. of ≢(3.10), with q:=¬(r≡q)>
			p ≢ ¬(q ≡ r) ≡ ¬(¬(p ≡ q) ≡ r)
		=    <Def. of ≢(3.10), with p,q:=¬(¬(q≡p),r>
			p ≢ ¬(q ≡ r) ≡ ¬(p ≡ q) ≢ r
		=    <Def. of ≢(3.10), with p,q:=q,r —replace ¬(q≡r) by q≢r>
			p ≢ q ≢ r ≡ ¬(p ≡ q) ≢ r
		=    <Def. of ≢(3.10), —replace ¬(p≡q) by p≢q>
			(p ≢ (q ≢ r)) ≡ ((p ≢ q) ≢ r)
3.13 Prove Mutual associativity (3.18), ((p ≢ q) ≡ r)≡(p ≢ (q ≡ r)), using the heuristic of
	 Definition elimination(3.23) —by eliminating ≢, using a property of ≡, and
	 reintroducing ≢.
	 Answer:
		((p ≢ q) ≡ r) ≡ (p ≢ (q ≡ r))
		=    <Def. of ≢(3.10) —replace (p≢q) by ¬(p≡q)>
			(¬(p ≡ q) ≡ r) ≡ (p ≢ (q ≡ r))
		=    <Def. of ≢(3.10) with q:=(q ≡ r) —replace p≢(q≡r) by ¬(p≡(q≡r))>
			(¬(p ≡ q) ≡ r) ≡ ¬(p ≡ (q ≡ r))
		=    <Symmetry of ≡(3.2), with p,q:=(¬(p≡q)≡r),¬(p≡(q≡r))>
			¬(p ≡ (q ≡ r)) ≡ (¬(p ≡ q) ≡ r)
		=    <Def. of ≢(3.10) with q:=(q ≡ r) —replace ¬(p ≡ (q ≡ r)) by p≢(q≡r)>
			p ≢ (q ≡ r) ≡ (¬(p ≡ q) ≡ r)
		=    <Def. of ≢(3.10) with p:=¬(p≡q) —replace ¬(p≡q) by p≢q>
			p ≢ (q ≡ r) ≡ p ≢ q ≡ r
3.14 Prove Mutual interchangeability(3.19), p ≢ q ≡ r ≡ p ≡ q ≢ r, using the heuristic
	 of Definition elimination(3.23) —by eliminating ≢, using a property of ≡, and
	 reintroducing ≢.
	 Answer:
		p ≢ q ≡ r ≡ p ≡ q ≢ r
		=    <Def. of ≢(3.10) —replace p≢q by ¬(p≡q)>
			¬(p ≡ q) ≡ r ≡ p ≡ q ≢ r
		=    <Def. of ≢(3.10), with p,q:=q,r —replace q≢r by ¬(q≡r)>
			¬(p ≡ q) ≡ r ≡ p ≡ ¬(q ≡ r)
		=    <Symmetry of ≡(3.2), with p,q:=¬(p≡q)≡r,p≡¬(q≡r)>
			p ≡ ¬(q ≡ r) ≡ ¬(p ≡ q) ≡ r
		=    <Def. of ≢(3.10) —replace ¬(q≡r) by q≢r>
			p ≡ q ≢ r ≡ ¬(p ≡ q) ≡ r
		=    <Def. of ≢(3.10) —replace ¬(p≡q) by p≢q>
			p ≡ q ≢ r ≡ p ≢ q ≡ r
3.15 Assume that ≡, ¬, true, and false have meanings as given in Sec. 2.1.
	 Show that axioms (3.24)-(3.28) uniquely determine operator ∨ —only one of the
	 operators of the truth table for binary operators on page 26 can be assigned to it.
	 Answer:
		OP = |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  T  |     |     |     |     |     |     |     |p=q≡q=p
		|3-25|  T  |     |     |     |     |     |     |     |(p=q)=r≡p=(q=r)
		|3-26|  T  |     |     |     |     |     |     |     |p=p≡p
		|3-27|  T  |     |     |     |     |     |     |     |p=(q≡r)≡p=q≡p=r
		|3-28|  F  |     |     |     |     |     |     |     |p=¬p
		------------------------------------------------------
		OP ≠ |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  F  |     |     |     |     |     |     |     |p≠q≡q≠p
		|3-25|     |     |     |     |     |     |     |     |(p≠q)≠r≡p≠(q≠r)
		|3-26|     |     |     |     |     |     |     |     |p≠p≡p
		|3-27|     |     |     |     |     |     |     |     |p≠(q≡r)≡p≠q≡p≠r
		|3-28|     |     |     |     |     |     |     |     |p≠¬p
		------------------------------------------------------
		OP ∨ |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  T  |  T  |  T  |  T  |  T  |  T  |  T  |  T  |p∨q≡q∨p
		|3-25|  T  |  T  |  T  |  T  |  T  |  T  |  T  |  T  |(p∨q)∨r≡p∨(q∨r)
		|3-26|  T  |  T  |  T  |  T  |  T  |  T  |  T  |  T  |p∨p≡p
		|3-27|  T  |  T  |  T  |  T  |  T  |  T  |  T  |  T  |p∨(q≡r)≡p∨q≡p∨r
		|3-28|  T  |  T  |  T  |  T  |  T  |  T  |  T  |  T  |p∨¬p
		------------------------------------------------------
		OP ∧ |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  T  |     |     |     |     |     |     |     |p∧q≡q∧p
		|3-25|  T  |     |     |     |     |     |     |     |(p∧q)∧r≡p∧(q∧r)
		|3-26|  T  |     |     |     |     |     |     |     |p∧p≡p
		|3-27|  T  |     |     |     |     |     |     |     |p∧(q≡r)≡p∧q≡p∧r
		|3-28|  F  |     |     |     |     |     |     |     |p∧¬p
		------------------------------------------------------
		OP → |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  T  |     |     |     |     |     |     |     |p→q≡q→p
		|3-25|  T  |     |     |     |     |     |     |     |(p→q)→r≡p→(q→r)
		|3-26|  T  |     |     |     |     |     |     |     |p→p≡p
		|3-27|  T  |     |     |     |     |     |     |     |p→(q≡r)≡p→q≡p→r
		|3-28|  F  |     |     |     |     |     |     |     |p→¬p
		------------------------------------------------------
		OP ⇐ |T,T,T|F,F,F|T,F,F|F,T,T|F,T,F|T,F,T|F,F,T|T,T,F|
		|3-24|  T  |  T  |     |     |     |     |     |     |p⇐q≡q⇐p
		|3-25|  T  |  F  |     |     |     |     |     |     |(p⇐q)⇐r≡p⇐(q⇐r)
		|3-26|  T  |     |     |     |     |     |     |     |p⇐p≡p
		|3-27|  T  |     |     |     |     |     |     |     |p⇐(q≡r)≡p⇐q≡p⇐r
		|3-28|  T  |     |     |     |     |     |     |     |p⇐¬p
		------------------------------------------------------
3.16 Prove that the zero of a binary operator ⊕ is unique. (An object is unique
	 if, when we assume that two of them B and C exist, we can prove B=C.)
	 Answer:
		p ≢ q ≡ (¬p ∧ q) ∨ (p ∧ ¬q)
			|t|t|f|
			|f|f|f|
			|t|f|t|
			|f|t|t| No zero
3.17 Prove Identity of ∨(3.30), p ∨ false ≡ p, by transforming its more structured
	 side into its simpler side. Theorem (3.15) may be a suitable way to introduce
	 an equivalence.
	 Answer:
		 p ∨ false
		 =    <Def. of false(3.8) —replace false by ¬true>
			p ∨ ¬true
		 =    <Identity of ≡(3.3), with q:=p —replace true by p≡p>
			p ∨ ¬(p ≡ p)
		 =    <Axiom, Distributivity of ¬ over ≡(3.9) and ∨ over ≡(3.27), with q:=p>
			p ∨ ¬p ≡ p ∨ p
		 =    <Excluded Middle(3.28) —replace p∨¬p by true>
			true ≡ p ∨ p
		 =    <Idempotency of ∨(3.26) —replace p∨p by p>
			true ≡ p
		 =    <Identity of ≡(3.3), with q:=p —replace true by p≡p>
			p ≡ p ≡ p
		 =    <Symmetry of ≡(3.2) with q:=p —replace p≡q≡q by p>
			p
		true
		=    <Identity of ≡(3.3), with q:=true —replace true by true≡true>
			true ≡ true
		=    <Identity of ≡(3.3) with q:=p —replace true by p≡p>
			true ≡ p ≡ p
		=    <Idempotency of ∨(3.26) —replace p by p∨p>
			true ≡ p ∨ p ≡ p
		=    <Excluded middle(3.28) —replace true by p∨¬p>
			p ∨ ¬p ≡ p ∨ p ≡ p
		=    <Distributivity of ∨ over ≡(3.27), with q:=p
			  —replace p∨¬p≡p∨p≡p by p∨(¬p≡p)>
			p ∨ (¬p ≡ p) ≡ p
		=    <Distributivity of ¬ over ≡(3.9), with q:=p —replace (¬p≡p) by ¬(p≡p)>
			p ∨ ¬(p ≡ p) ≡ p
		=    <Reflexivity of ≡(3.5) —replace p≡p by true>
			p ∨ ¬true ≡ p
		=    <Def. of false(3.8) —replace ¬true by false>
			p ∨ false ≡ p
3.18 Prove distributivity of ∨ over ∨(3.31), p ∨ (q ∨ r) ≡ (p ∨ q) ∨ (p ∨ r)
	 This proof requires only the symmetry, associativity, and idempotency of ∨.
	 Answer:
		(p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24) —replace p∨q by q∨p>
			(q ∨ p) ∨ (p ∨ r)
		=    <Associativity of ∨(3.25), with p,q:=(q ∨ p),p>
			(q ∨ (p ∨ r)) ∨ p
		=    <Symmetry of ∨(3.24), with q:=r —replace p∨r by r∨p>
			(q ∨ (r ∨ p)) ∨ p
		=    <Associativity of ∨(3.25), with p,q,r:=q,r,p>
			((q ∨ r) ∨ p) ∨ p
		=    <Associativity of ∨(3.25), with p,q,r:=(q ∨ r),p,p>
			(q ∨ r) ∨ (p ∨ p)
		=    <Symmetry of ∨(3.24) ,with p,q:=(q∨r),(p∨p) —replace (q∨r)∨(p∨p) by (p∨p)∨(q∨r)>
			(p ∨ p) ∨ (q ∨ r)
		=    <Idempotency of ∨(3.26) —replace (p∨p) by p>
			p ∨ (q ∨ r)
3.19 Prove theorem(3.32), p ∨ q ≡ p ∨ ¬q ≡ p. Note that the pattern p ∨ q ≡ p ∨ ¬q
	 matches the RHS of distributivity axiom(3.27), with r:=¬q,
	 so consider transforming p ∨ q ≡ p ∨ ¬q to p.
	 Answer:
		p ∨ q ≡ p ∨ ¬q
		=    <Distributivity of ∨ over ≡(3.27), with r:=¬q>
			p ∨ (q ≡ ¬q)
		=    <Symmetry of ≡(3.2), with p,q:=q,¬q>
			p ∨ (¬q ≡ q)
		=    <Distributivity of ¬ over ≡(3.9), with p:=q>
			p ∨ ¬(q ≡ q)
		=    <Identity of≡(3.3) —replace q≡q by true>
			p ∨ ¬true
		=    <Def. of false(3.8) —replace ¬true by false>
			p ∨ false
		=    <Identity of ∨(3.30) —replace p∨false by p>
			p
3.20 Show the validity of the Golden rule, (3.35), by constructing a truth table for it.
	 Answer:
		|p|q|p∧q|p∨q|(p≡q)≡(p∧q≡p∨q)|
		|t|t| t | t |       t       |
		|f|f| f | f |       t       |
		|t|f| f | t |       t       |
		|f|t| f | t |       t       |
3.21 Prove that the only distinct formulas (up to interchanging p and q) involving
	 variables p, q, ≡, and ∨ are: p, p ≡ p, p ≡ q, p ∨ q, p ∨ q ≡ q, and p ≡ q ≡ p ∨ q.
	 Answer:
		SKIP
3.22 Prove Symmetry of ∧(3.36), p ∧ q ≡ q ∧ p, using the heuristic of Definition
	 Elimination(3.23) —eliminate ∧ (using its definition, the Golden rule),
	 manipulate, and then reintroduce ∧.
	 Answer:
		p ∧ q ≡ q ∧ p
		=    <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
			p ≡ q ≡ p ∨ q ≡ q ∧ p
		=    <Golden rule(3.35) with p,q:=q,p —replace q∧p by q≡p≡q∨p>
			p ≡ q ≡ p ∨ q ≡ q ≡ p ≡ q ∨ p
		=    <Symmetry of ≡(3.2) with p,q:=p≡q≡p∨q,q≡p≡q∨p
			  —replace p≡q≡p∨q≡q≡p≡q∨p by q≡p≡q∨p≡p≡q≡p∨q>
			q ≡ p ≡ q ∨ p ≡ p ≡ q ≡ p ∨ q
		=    <Golden rule(3.35) with p,q:=q,p —replace q≡p≡q∨p by q∧p>
			q ∧ p ≡ p ≡ q ≡ p ∨ q
		=    <Golden rule(3.35) with —replace p≡q≡p∨q by p∧q>
			q ∧ p ≡ p ∧ q
3.23 Prove Idempotency of ∧(3.38), p ∧ p ≡ p, using the heuristic of Definition
	 elimination (3.23) —eliminate ∧ (using its definition, the Golden rule) and manipulate.
	 Answer:
		p ∧ p
		=    <Golden rule(3.38), with q:=p —replace p∧p by p≡p≡p∨p>
			p ≡ p ≡ p ∨ p
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ p ≡ p
		=    <Symmetry of ≡(3.2) with q:=p —replace p≡q≡q by p>
			p
3.24 Prove Zero of ∧(3.40), p ∧ false ≡ false, using the heuristic of definition
	 elimination(3.23) —eliminate ∧ (using its definition, the Golden rule) and
	 manipulate.
	 Answer:
		p ∧ false
		=    <Golden rule(3.35) with q:=false —replace p∧false by p≡false≡p∨false>
			p ≡ false ≡ p ∨ false
		=    <Identity of ∨(3.30) —replace p∨false by p>
			p ≡ false ≡ p
		=    <Symmetry of ≡(3.2), with q:=false —replace p≡false by false≡p>
			false ≡ p ≡ p
		=    <Def. of false(3.8) —replace false by ¬true>
			¬true ≡ p ≡ p
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=true,p
			  —replace ¬true≡p by ¬(true≡p)>
			¬(true ≡ p) ≡ p
		=    <Associativity of ≡(3.1), with p,q,r:=true,p,p
			  —replace ¬(true≡p)≡p by ¬true≡(p≡p)>
			¬true ≡ (p ≡ p)
		=    <Identity of ≡(3.3), with q:=p —replace p≡p by true>
			¬true ≡ true
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=true,true
			  —replace ¬true≡true by ¬(true≡true)>
			¬(true≡true)
		=    <Identity of ≡(3.3), q:=true —replace true≡true by true>
			¬true
		=    <Def. of false(3.8) —replace ¬true by false>
			false
3.25 Prove Distributivity of ∧ over ∧(3.41), p ∧ (q ∧ r) ≡ (p ∧ q) ∧ (p ∧ r)
	 Answer:
		p ∧ (q ∧ r)
		=    <Idempotency of ∧(3.38) —replace p by p∧p>
			p ∧ p ∧ (q ∧ r)
		=    <Associativity of ∧(3.37) —replace p∧(q∧r) by (p∧q)∧r>
			p ∧ (p ∧ q) ∧ r
		=    <Symmetry of ∧(3.36), with p,q:=p∧(p∧q),r —replace p∧(p∧q)∧r
			  by r∧p∧(p∧q)>
			r ∧ p ∧ (p ∧ q)
		=    <Symmetry of ∧(3.36), with p,q:=r,p —replace r∧p by p∧r>
			p ∧ r ∧ (p ∧ q)
3.26 Prove Contradiction(3.42), p ∧ ¬p ≡ false, using the heuristic of Definition
	 elimination(3.23) —eliminate ∧ (using its definition, the Golden rule) and
	 manipulate.
	 Answer:
		p ∧ ¬p
		=    <Golden rule(3.35), with q:=¬p —replace p∧¬p by p≡¬p≡p∨¬p>
			p ≡ ¬p ≡ p ∨ ¬p
		=    <Excluded middle(3.28) —replace p∨¬p by true>
			p ≡ ¬p ≡ true
		=    <Identity of ≡(3.3), with q:=p —replace true by p≡p>
			p ≡ ¬p ≡ p ≡ p
		=    <Symmetry of ≡(3.2), with q:=¬p —replace p≡¬p by ¬p≡p>
			¬p ≡ p ≡ p ≡ p
		=    <Distributivity of ¬ over ≡(3.9) —replace ¬p≡p by ¬(p≡p)>
			¬(p ≡ p) ≡ p ≡ p
		=    <Identity of ≡(3.3), with q:=p —replace p≡p by true>
			¬true ≡ p ≡ p
		=    <Def. of false(3.8) —replace ¬true by false>
			false ≡ p ≡ p
		=    <Identity of ∨(3.30) —replace p by p∨false>
			false ≡ p ≡ p ∨ false
		=    <Symmetry of ≡(3.2),with p,q:=false,p —replace false≡p by p≡false>
			p ≡ false ≡ p ∨ false
		=    <Golden rule(3.35), with q:=false —replace p≡false≡p∨false
			  by p ∧ false>
			p ∧ false
		=    <Zero of ∧(3.40) —replace p∧false by false>
			false
3.27 Prove Absorption(3.43a), p ∧ (p ∨ q) ≡ p, using the heuristic of Definition
	 elimination(3.23) —eliminate ∧ (using its definition, the Golden rule) and
	 manipulate.
	 Answer:
		p ∧ (p ∨ q)
		=    <Golden rule(3.35), with q:=(p∨q) —replace p∧(p∨q) by p≡p∨q≡p∨p∨q>
			p ≡ p ∨ q ≡ p ∨ p ∨ q
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ p ∨ q ≡ p ∨ q
		=    <Symmetry of ≡(3.32), with q:=p∨q —replace p≡p∨q≡p∨q by p>
			p
3.28 Prove Absorption(3.43b), p ∨ (p ∧ q) ≡ p. Use the golden rule.
	Answer:
		p ∨ (p ∧ q)
		=    <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
			p ∨ (p ≡ q ≡ p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), with q,r:=(p≡q),(p∨q)
			  —replace p∨(p≡q≡p∨q) by p∨(p≡q)≡p∨(p∨q)>
			p ∨ p ∨ (p ≡ q) ≡ p ∨ (p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), —replace p∨(p≡q) by p∨p≡p∨q
			p ∨ p ∨ p ≡ p ∨ q ≡ p ∨ (p ∨ q)
		=    <Associativity of ∨(3.25), with q,r:=p,q —replace p∨(p∨q)
			  by (p∨p)∨q)
			p ∨ p ∨ p ≡ p ∨ q ≡ (p ∨ p) ∨ q
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ∨ p ≡ p ∨ q ≡ (p ∨ p) ∨ q
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ∨ p ≡ p ∨ q ≡ p ∨ q
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ p ∨ q ≡ p ∨ q
		=    <Symmetry of ∨(3.24), with q:=p∨q —replace p≡p∨q≡p∨q by p>
			p
3.29 Prove Absorption(3.44b), p ∨ (¬p ∧ q) ≡ p ∨ q. Use the Golden rule and manipulate.
	 Answer:
	 p ∨ (¬p ∧ q)
	 =    <Distributivity of ∨ over ∧(3.45), with q,r:=¬p,q —replace p∨(¬p∧q)
		   by p∨¬p∧p∨q>
			p ∨ ¬p ∧ p ∨ q
	 =    <Golden rules(3.35), with p,q:=(p∨¬p),(p∨q) —replace p∨¬p∧p∨q by
		   p∨¬p≡p∨q≡p∨¬p∨p∨q>
			p ∨ ¬p ≡ p ∨ q ≡ p ∨ ¬p ∨ p ∨ q
	 =    <Metatheorem(3.7), Excluded middle(3.28)
		   and Reflexivity of ≡(3.5) —replace p∨¬p by (p=p)>
			(p ≡ p) ≡ p ∨ q ≡ (p ≡ p) ∨ p ∨ q
	 =    <Golden rules(3.35), with p,q:=(p≡p),(p∨q) —replace (p≡p)≡p∨q≡true∨p∨q
		   by (p≡p)∧p∨q>
			(p ≡ p) ∧ p ∨ q
	 =    <Symmetry of ∧(3.36), with p,q:=(p≡p),p∨q —replace (p≡p)∧p∨q by p∨q∧true>
			p ∨ q ∧ (p ≡ p)
	 =    <Metatheorem(3.7), Reflexivity of ≡(3.5) and (3.4) —replace p≡p by true
		   and Identity of ∧(3.39) with p:=p∨q  —replace p∨q∧true by p∨q>
			p ∨ q
3.30 Prove Distributivity of ∨ over ∧(3.45), p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r),
	 using the heuristic of Definition elimination(3.23) —eliminate ∧ (using its
	 definition, the Golden rule), manipulate, and reintroduce ∧ using the Golden
	 rule again.
	 Answer:
		p ∨ (q ∧ r)
	 =    <Golden rule(3.35), with p,q:=q,r —replace q∧r by q≡r≡q∨r>
		p ∨ (q ≡ r ≡ q ∨ r)
	 =    <Distributivity of ∨ over ≡(3.27) with q,r:=q≡r,q∨r —replace p∨(q≡r≡q∨r)
		   by p∨(q≡r)≡p∨(q∨r)>
		p ∨ (q ≡ r) ≡ p ∨ (q ∨ r)
	 =    <Distributivity of ∨ over ≡(3.27) —replace p∨(q≡r) by p∨q≡p∨r>
		p ∨ q ≡ p ∨ r ≡ p ∨ (q ∨ r)
	 =    <Distributivity of ∨ over ∨(3.31), -replace p∨(q∨r) by p∨q∨r>
		p ∨ q ≡ p ∨ r ≡ p ∨ q ∨ r
	 =    <Golden rule(3.35), with p,q:=p∨q,p∨r —replace p∨q≡p∨r≡p∨q∨p∨r
		   by (p∨q)∧(p∨r)>
		(p ∨ q) ∧ (p ∨ r)
3.31 Prove Distributivity of ∧ over ∨(3.46). It cannot be proved in the same manner
	 as Distributivity of ∨ over ∧(3.45) because ∧ does not distribute over ≡ so
	 nicely. Instead, prove it using (3.45) and Absorption.
	 Answer:
		(p ∧ q) ∨ (p ∧ r)
		=    <Distributivity of ∨ over ∧(3.45), with p,q:=(p∧q),p —replace
			  (p∧q)∨(p∧r) by ((p∧q)∨p)∧((p∧q)∨r)>
			((p ∧ q) ∨ p) ∧ ((p ∧ q) ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=(p∧q),p —replace (p∧q)∨p by p∨(p∧q)>
			(p ∨ (p ∧ q)) ∧ ((p ∧ q) ∨ r)
		=    <Absorption(3.43b) —replace p∨(p∧q) by p>
			p ∧ ((p ∧ q) ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=(p∧q),r —replace (p∧q)∨r by r∨(p∧q)>
			p ∧ (r ∨ (p ∧ q))
		=    <Distributivity of ∨ over ∧(3.45), with p,q,r:=r,p,q —replace
			  r∨(p∧q) by (r∨p)∧(r∨q)>
			p ∧ ((r ∨ p) ∧ (r ∨ q))
		=    <Associativity of ∧, with q,r:=(r∨p),(r∨q) —replace p∧((r∨p)∧(r∨q))
			  by (p∧(r∨p))∧(r∨q)>
			(p ∧ (r ∨ p)) ∧ (r ∨ q)
		=    <Symmetry of ∨(3.24), with p,q:=r,p —replace r∨p by p∨r>
			(p ∧ (p ∨ r)) ∧ (r ∨ q)
		=    <Absorption(3.43b), with q:=r —replace p∧(p∨r) by p>
			p ∧ (r ∨ q)
		=    <Symmetry of ∨(3.24), with p,q:=q,r —replace r∨q by q∨r>
			p ∧ (q ∨ r)
	 GOLDEN RULE(LOWLEVEL HARDCORE):
		(p ∧ q) ∨ (p ∧ r)
		=    <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
			(p ≡ q ≡ p ∨ q) ∨ (p ∧ r)
		=    <Golden rule(3.35) —replace p∧r by p≡r≡p∨r>
			(p ≡ q ≡ p ∨ q) ∨ (p ≡ r ≡ p ∨ r)
		=    <Distributivity of ∨ over ≡, with p,q,r:=(p≡q≡p∨q),(p≡r),(p∨r)
			  —replace (p≡q≡p∨q)∨(p≡r≡p∨r) by (p≡q≡p∨q)∨(p≡r)≡(p≡q≡p∨q)∨(p∨r)>
			(p ≡ q ≡ p ∨ q) ∨ (p ≡ r) ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Distributivity of ∨ over ≡, with p,q,r:=(p≡q≡p∨q),p,r
			  —replace (p≡q≡p∨q)∨(p≡r) by ((p≡q≡p∨q)∨p)≡((p≡q≡p∨q)∨r)>
			(p ≡ q ≡ p ∨ q) ∨ p ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=(p≡q≡p∨q),p —replace (p≡q≡p∨q)∨p
			  by p∨(p≡q≡p∨q)>
			p ∨ (p ≡ q ≡ p ∨ q) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Distributivity of ∨ over ≡(3.27), with q,r:=(p≡q),(p∨q) —replace
			  p∨(p≡q≡p∨q) by (p∨(p≡q))≡(p∨(p∨q))>
			p ∨ (p ≡ q) ≡ p ∨ (p ∨ q) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Distributivity of ∨ over ≡(3.27), with q,r:=p,q —replace p∨(p≡q) by
			  p∨p≡p∨q>
			p ∨ p ≡ p ∨ q ≡ p ∨ (p ∨ q) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24) —replace p∨q by q∨p>
			p ∨ p ≡ p ∨ q ≡ p ∨ (q ∨ p) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with q:=(q∨p) —replace p∨(q∨p) by (q∨p)∨p>
			p ∨ p ≡ p ∨ q ≡ (q ∨ p) ∨ p ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Associativity of ∨ over ∨(3.25), with p,q,r:=q,p,p —replace (q∨p)∨p by
			  q∨(p∨p)>
			p ∨ p ≡ p ∨ q ≡ q ∨ (p ∨ p) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ p ∨ q ≡ q ∨ (p ∨ p) ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ p ∨ q ≡ q ∨ p ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=q,p —replace q∨p by p∨q>
			p ≡ p ∨ q ≡ p ∨ q ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with q:=p∨q —replace p≡p∨q≡p∨q by p>
			p ≡ (p ≡ q ≡ p ∨ q) ∨ r ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=(p≡q≡p∨q),r —replace (p≡q≡p∨q)∨r
			  by r∨(p≡q≡p∨q)>
			p ≡ r ∨ (p ≡ q ≡ p ∨ q) ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,(p≡q),(p∨q) —replace 
			  r∨(p≡q≡p∨q) by (r∨(p≡q))≡(r∨(p∨q))>
			p ≡ r ∨ (p ≡ q) ≡ r ∨ p ∨ q ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p,q —replace r∨(p≡q)
			  by (r∨p)≡(r∨q)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ (p ≡ q ≡ p ∨ q) ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=(p≡q≡p∨q),(p∨r) —replace (p≡q≡p∨q)∨(p∨r)
			  by (p∨r)∨(p≡q≡p∨q)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ (p ∨ r) ∨ (p ≡ q ≡ p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=(p∨r),(p≡q),(p∨q) —replace
			  (p∨r)∨(p≡q≡p∨q) by (p∨r)∨(p≡q)≡(p∨r)∨(p∨q)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ (p ∨ r) ∨ (p ≡ q) ≡ (p ∨ r) ∨ (p ∨ q)
		=    <Distributivity of ∨ over ≡(3.27), with p,q,r:=(p∨r),p,q —replace
			  (p∨r)∨(p≡q) by (p∨r)∨p≡(p∨r)∨q>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ (p ∨ r) ∨ p ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ (p ∨ q)
		=    <Symmetry of ∨(3.24), with q:=r —replace p∨r by r∨p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ (r ∨ p) ∨ p ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ (p ∨ q)
		=    <Associativity of ∨(3.25), with p,q,r:=r,p,p —replace (r∨p)∨p by r∨(p∨p)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ (p ∨ p) ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ (p ∨ q)
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ (p ∨ q)
		=    <Distributivity of ∨ over ∨(3.31), with p,q,r:=(p∨r),p,q —replace
			  (p∨r)∨(p∨q) by ((p ∨ r) ∨ p) ∨ ((p ∨ r) ∨ q)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ ((p ∨ r) ∨ p) ∨ ((p ∨ r) ∨ q)
		=    <Symmetry of ∨(3.24), with q:=r —replace p∨r by r∨p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ ((r ∨ p) ∨ p) ∨ ((p ∨ r) ∨ q)
		=    <Associativity of ∨(3.25), with p,q,r:=r,p,p —replace (r∨p)∨p
			  by r∨(p∨p)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (r ∨ (p ∨ p)) ∨ ((p ∨ r) ∨ q)
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (r ∨ p) ∨ ((p ∨ r) ∨ q)
		=    <Symmetry of ∨(3.24), with p,q:=r,p —replace r∨p by p∨r>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ ((p ∨ r) ∨ q)
		=    <Symmetry of ∨(3.24), with p,q:=(p∨r),q —replace (p∨r)∨q by q ∨ (p ∨ r)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (p ∨ r) ∨ (q ∨ (p ∨ r))
		=    <Symmetry of ∨(3.24), with p,q:=(p∨r),(q∨(p∨r)) —replace (p∨r)∨(q∨(p∨r))
			  by (q ∨ (p ∨ r)) ∨ (p ∨ r)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ (q ∨ (p ∨ r)) ∨ (p ∨ r)
		=    <Associativity of ∨(3.25), with p,q:=(q∨(p∨r)),p —replace (q∨(p∨r))∨(p∨r)
			  by q∨((p∨r)∨(p∨r))>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((p ∨ r) ∨ (p ∨ r))
		=    <Distributivity of ∨ over ∨(3.31), with p,q,r:=(p∨r),p,r —replace (p∨r)∨(p∨r)
			  by ((p ∨ r) ∨ p) ∨ ((p ∨ r) ∨ r)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ (((p ∨ r) ∨ p) ∨ ((p ∨ r) ∨ r))
		=    <Symmetry of ∨(3.24), with q:=r —replace p∨r by r∨p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ (((r ∨ p) ∨ p) ∨ ((p ∨ r) ∨ r))
		=    <Associativity of ∨(3.25), with p,q,r:=r,p,p —replace (r∨p)∨p by r∨(p∨p)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((r ∨ (p ∨ p)) ∨ ((p ∨ r) ∨ r))
		=    <Idempotency of ∨(3.26) —replace p∨p by p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((r ∨ p) ∨ ((p ∨ r) ∨ r))
		=    <Associativity of ∨(3.25), with q:=r —replace (p∨r)∨r by p ∨ (r ∨ r)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((r ∨ p) ∨ (p ∨ (r ∨ r)))
		=    <Idempotency of ∨(3.26), with p:=r —replace r∨r by r>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((r ∨ p) ∨ (p ∨ r))
		=    <Symmetry of ∨(3.24), with p,q:=r,p —replace r∨p by p∨r>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ ((p ∨ r) ∨ (p ∨ r))
		=    <Idempotency of ∨(3.26), with p:=(p∨r) —replace (p∨r)∨(p∨r) by (p∨r)>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ (p ∨ r) ∨ q ≡ q ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p=(p∨r) —replace (p∨r)∨q by q∨(p∨r )>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p ≡ q ∨ (p ∨ r) ≡ q ∨ (p ∨ r)
		=    <Symmetry of ∨(3.24), with p,q:=r∨p,q∨(p∨r) —replace r∨p≡q∨(p∨r)≡q∨(p∨r)
			  by r∨p>
			p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q ≡ r ∨ p
		=    <Symmetry of ≡(3.2), with p,q:=(p≡r∨p≡r∨q≡r∨(p∨q)),r∨p —replace
			  p≡r∨p≡r∨q≡r∨(p∨q)≡r∨p by r∨p≡p≡r∨p≡r∨q≡r∨(p∨q)>
			r ∨ p ≡ p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q
		=    <Symmetry of ≡(3.2), with p,q:=r∨p,p —replace r∨p≡p by p≡r∨p>
			p ≡ r ∨ p ≡ r ∨ p ≡ r ∨ q ≡ r ∨ p ∨ q
		=    <Symmetry of ≡(3.2), with q:=r∨p —replace p≡r∨p≡r∨p by p>
			p ≡ r ∨ q ≡ r ∨ p ∨ q
		=    <Symmetry of ∨(3.24), with p:=r —replace r∨q by q∨r>
			p ≡ q ∨ r ≡ r ∨ p ∨ q
		=    <Symmetry of ∨(3.24), with p,q:=r,p —replace r∨p by p∨r>
			p ≡ q ∨ r ≡ p ∨ r ∨ q
		=    <Symmetry of ∨(3.24), with p:=r —replace r∨q by q∨r>
			p ≡ q ∨ r ≡ p ∨ q ∨ r
		=    <Golden rule(3.35), with q:=q∨r —replace p≡q∨r≡p∨q∨r by p∧(q∨r)>
			p ∧ (q ∨ r)
3.47 De Morgan:
		1. ¬(p ∧ q) ≡ ¬ p ∨ ¬q
		2. ¬(p ∨ q) ≡ ¬ p ∧ ¬q
3.32 Prove De Morgan(3.47a), ¬(p ∧ q) ≡ ¬p ∨ ¬q. Start by using the Golden rule;
	 (3.32) should come in handy.
	 Answer:
		¬(p ∧ q)
		=    <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
			¬(p ≡ q ≡ p ∨ q)
		=    <(3.32) —replace p∨q by p∨¬q≡p>
			¬(p ≡ q ≡ p ∨ ¬q ≡ p)
		=    <Symmetry of ≡(3.2), with p,q:=p∨¬q,p≡q —replace p≡q≡p∨¬q
			  by p∨¬q≡p≡q>
			¬(p ∨ ¬q ≡ p ≡ q ≡ p)
		=    <Symmetry of ≡(3.2) —replace p≡q by q≡p>
			¬(p ∨ ¬q ≡ q ≡ p ≡ p)
		=    <Symmetry of ≡(3.2), with p,q:=q,p —replace q≡p≡p by q>
			¬(p ∨ ¬q ≡ q)
		=    <Symmetry of ≡(3.2), with p,q:=p∨¬q,q —replace p∨¬q≡q by q≡p∨¬q>
			¬(q ≡ p ∨ ¬q)
		=    <Distributivity of ¬ over ≡(3.9), with p,q:=q,p∨¬q —replace ¬(q≡p∨¬q)
			  ¬q≡p∨¬q>
			¬q ≡ p ∨ ¬q
		=    <Symmetry of ∨(3.24), with q:=¬q —replace p∨¬q by ¬q∨p>
			¬q ≡ ¬q ∨ p
		=    <(3.32), with p,q:=¬q,p —replace ¬q≡¬q∨p by ¬q∨¬p>
			¬q ∨ ¬p
		=    <Symmetry of ∨(3.24), with p,q:=¬q∨¬p —replace ¬q∨¬p by ¬p∨¬q>
			¬p ∨ ¬q
3.33 Prove De Morgan(3.47b), ¬(p ∨ q) ≡ ¬p ∧ ¬q, beginning with the LHS and using
	 the Golden rule.
	 Answer:
	 ¬(p ∨ q)
	 =   <(3.32) —replace p∨q by p≡p∨¬q>
		¬(p ≡ p ∨ ¬q)
	 =  <(3.32) —replace p by p∨q≡p∨¬q>
		¬(p ∨ q ≡ p ∨ ¬q ≡ p ∨ ¬q)
	 =  <(3.32) —replace p∨q by p≡p∨¬q>
		¬(p ≡ p ∨ ¬q ≡ p ∨ ¬q ≡ p ∨ ¬q)
	 =   <Distributivity of ¬ over≡(3.9), with p,q:=p,p∨¬q≡p∨¬q≡p∨¬q —replace
		  ¬(p≡p∨¬q≡p∨¬q≡p∨¬q) by ¬p≡p∨¬q≡p∨¬q≡p∨¬q>
		¬p ≡ p ∨ ¬q ≡ p ∨ ¬q ≡ p ∨ ¬q
	 =   <Idempotency of ∨(3.26) —replace p by p∨p>
		¬p ≡ p ∨ ¬q ≡ p ∨ p ∨ ¬q ≡ p ∨ ¬q
	 =   <Symmetry of ∨(3.26), with q:=¬q —replace p∨¬q by ¬q∨p>
		¬p ≡ p ∨ ¬q ≡ p ∨ ¬q ∨ p ≡ p ∨ ¬q
	 =   <(3.32), with p,q:=p∨¬q,p —replace p∨¬q∨p≡p∨¬q by p∨¬q∨¬p>
		¬p ≡ p ∨ ¬q ≡ p ∨ ¬q ∨ ¬p
	 =   <Symmetry of ∨(3.26), with p,q:=p∨¬q,¬p —replace p∨¬q∨¬p by ¬p∨p∨¬q>
		¬p ≡ p ∨ ¬q ≡ ¬p ∨ p ∨ ¬q
	 =   <Golden rule(3.35), with p,q:=¬p,p∨¬q —replace ¬p≡p∨¬q≡¬p∨p∨¬q
		  by ¬p∧(p∨¬q)>
		¬p ∧ (p ∨ ¬q)
	 =   <Double negation(3.12) —replace p by ¬¬p>
		¬p ∧ (¬¬p ∨ ¬q)
	 =   <Absorption(3.44a), with p,q:=¬p,¬q —replace ¬p∧(¬¬p∨¬q) by ¬p∧¬q>
		¬p ∧ ¬q
3.34 Prove (p ∧ q) ∨ (p ∧ ¬q) ≡ p
	 Answer:
	 (p ∧ q) ∨ (p ∧ ¬q)
	 =   <Distributivity of ∧ over ∨(3.46), with r:=¬q —replace (p∧q)∨(p∧¬q)
		  by p ∧(q ∨¬q)>
		p ∧ (q ∨ ¬q)
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=q and theorem(3.4)
		  —replace q∨¬q by true>
		p ∧ true
	 =   <Identity of ∧(3.39), —replace p∧true by p>
		p
3.35 Prove (3.48), p ∧ q ≡ p ∧ ¬q ≡ ¬p. Theorem(3.32) should come in
	 handy.
	 Answer:
	 p ∧ q
	 =   <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
		p ≡ q ≡ p ∨ q
	 =   <Symmetry of ∨(3.24) —replace p≡q by q≡p>
		q ≡ p ≡ p ∨ q
	 =   <(3.32) —replace p≡p∨q by p∨¬q>
		q ≡ p ∨ ¬q
	 =   <Golden rule(3.35), with q:=¬q —replace p∨¬q by p≡¬q≡p∧¬q>
		q ≡ p ≡ ¬q ≡ p ∧ ¬q
	 =   <Symmetry of ≡(3.2), with p,q:=q≡p,¬q —replace q≡p≡¬q by ¬q≡q≡p>
		¬q ≡ q ≡ p ≡ p ∧ ¬q
	 =   <Distributivity of ¬ over ≡(3.9), with p,q:=q,q≡p≡p∧¬q —replace ¬q≡q≡p≡p∧¬q
		  by ¬(q≡q≡p≡p∧¬q)>
		¬(q ≡ q ≡ p ≡ p ∧ ¬q)
	 =   <Symmetry of ≡(3.2), with p,q:=q≡q,p —replace q≡q≡p by p≡q≡q>
		¬(p ≡ q ≡ q ≡ p ∧ ¬q)
	 =   <Symmetry of ≡(3.2) —replace p≡q≡q by p>
		¬(p ≡ p ∧ ¬q)
	 =   <Distributivity of ¬ over ≡(3.9), with q:=p∧¬q —replace ¬(p≡p∧¬q)
		  by ¬p≡p∧¬q>
		¬p ≡ p ∧ ¬q
	 =   <Symmetry of ≡(3.2), with p,q:=¬p,p∧¬q —replace ¬p≡p∧¬q by p∧¬q≡¬p>
		p ∧ ¬q ≡ ¬p
3.36 Prove (3.50), p ∧ (q ≡ p) ≡ p ∧ q, using (3.49) with the instantiation
	 r:=p.
	 Answer:
	 p ∧ (q ≡ p)
	 =   <(3.49), with r:=p —replace p∧(q≡p) by p∧q≡p∧p≡p>
		p ∧ q ≡ p ∧ p ≡ p
	 =   <Identity of ∧(3.39) —replace p∧p by p>
		p ∧ q ≡ p ≡ p
	 =   <Symmetry of ≡(3.2), with p,q:=p∧q,p —replace p∧q≡p≡p by p ∧ q>
		p ∧ q
3.37 Prove Replacement(3.51), (p ≡ q) ∧ (r ≡ p) ≡ (p ≡ q) ∧ (r ≡ q), by proving
	 that the LHS and the RHS each equivale p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p.
	 The transformation of the LHS (or the RHS) to this expression can be done
	 by applying (3.27) three times.
	 Answer:
	 (p ≡ q) ∧ (r ≡ p) ≡ (p ≡ q) ∧ (r ≡ q)
	 =   <Golden rule(3.35), with p,q:=(p≡q),(r≡p) —replace (p≡q)∧(r≡p)
		  by (p≡q)≡(r≡p)≡(p≡q)∨(r≡p) twice>
		p ≡ q ≡ r ≡ p ≡ (p ≡ q) ∨ (r ≡ p) ≡ (p ≡ q) ≡ (r ≡ p) ≡ (p ≡ q) ∨ (r ≡ p)
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=(p≡q),r,p —replace (p≡q)∨(r≡p)
		  by ((p≡q)∨r)≡((p≡q)∨p) twice>
		p ≡ q ≡ r ≡ p ≡ ((p ≡ q) ∨ r) ≡ ((p ≡ q) ∨ p) ≡ p ≡ q ≡ r ≡ p ≡ ((p ≡ q) ∨ r) ≡ ((p ≡ q) ∨ p)
	 =   <Symmetry of ∨(3.24), with p,q:=(p≡q),r —replace (p≡q)∨r by r∨(p≡q) twice>
		p ≡ q ≡ r ≡ p ≡ (r ∨ (p ≡ q)) ≡ ((p ≡ q) ∨ p) ≡ p ≡ q ≡ r ≡ p ≡ (r ∨ (p ≡ q)) ≡ ((p ≡ q) ∨ p)
	 =   <Symmetry of ∨(3.24), with p,q:=(p≡q),p —replace (p≡q)∨p by p∨(p≡q) twice>
		p ≡ q ≡ r ≡ p ≡ (r ∨ (p ≡ q)) ≡ (p ∨ (p ≡ q)) ≡ p ≡ q ≡ r ≡ p ≡ (r ∨ (p ≡ q)) ≡ (p ∨ (p ≡ q))
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p,q —replace r∨(p≡q) by r∨p≡r∨q twice>
		p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ (p ∨ (p ≡ q)) ≡ p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ (p ∨ (p ≡ q))
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=p,p,q —replace p∨(p≡q) by p∨p≡p∨q twice>
		p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ∨ p ≡ p ∨ q ≡ p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ∨ p ≡ p ∨ q
	 =   <Idempotency of ∨(3.26) —replace p∨p by p twice>
		p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ≡ p ∨ q ≡ p ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=r∨p≡r∨q,p —replace r∨p≡r∨q≡p by p≡r∨p≡r∨q twice>
		p ≡ q ≡ r ≡ p ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ∨ q ≡ p ≡ q ≡ r ≡ p ≡ p ≡ r ∨ p ≡ r ∨ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=p≡q≡r,p —replace p≡q≡r≡p≡p by p≡q≡r twice>
			p ≡ q ≡ r ≡ r ∨ p ≡ r ∨ q ≡ p ∨ q ≡ p ≡ q ≡ r ≡ r ∨ p ≡ r ∨ q ≡ p ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=q,r —replace r∨q by q∨r twice>
			p ≡ q ≡ r ≡ r ∨ p ≡ q ∨ r ≡ p ∨ q ≡ p ≡ q ≡ r ≡ r ∨ p ≡ q ∨ r ≡ p ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=r∨p≡q∨r,p∨q —replace r∨p≡q∨r≡p∨q
		  by p∨q≡r∨p≡q∨r twice>
		p ≡ q ≡ r ≡ p ∨ q ≡ r ∨ p ≡ q ∨ r ≡ p ≡ q ≡ r ≡ p ∨ q ≡ r ∨ p ≡ q ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=r∨p,q∨r —replace r∨p≡q∨r by q∨r≡r∨p twice>
		p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p ≡ p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p
3.38 Prove Replacement(3.51), (p ≡ q) ∧ (r ≡ p) ≡ (p ≡ q) ∧ (r ≡ q), by making
	 immediate use of Distributivity of ∧ over ≡(3.49) to replace both equivalents.
	 Answer:
	 (p ≡ q) ∧ (r ≡ p) ≡ (p ≡ q) ∧ (r ≡ q)
	 =   <Distributivity of ∧ over ≡(3.49), with p,q,r:=(p≡q),r,p —replace (p≡q)∧(r≡p)
		  by ((p≡q)∧r)≡((p≡q)∧p)≡(p≡q) twice>
		((p ≡ q) ∧ r) ≡ ((p ≡ q) ∧ p) ≡ p ≡ q ≡ ((p ≡ q) ∧ r) ≡ ((p ≡ q) ∧ p) ≡ p ≡ q
	 =   <Symmetry of ∧(3.36), with p,q:=(p≡q),r —replace (p≡q)∧r by r∧(p≡q) twice>
		(r ∧ (p ≡ q)) ≡ ((p ≡ q) ∧ p) ≡ p ≡ q ≡ (r ∧ (p ≡ q)) ≡ ((p ≡ q) ∧ p) ≡ p ≡ q
	 =   <Symmetry of ∧(3.36), with p,q:=(p≡q),p —replace (p≡q)∧p by p∧(p≡q) twice>
		(r ∧ (p ≡ q)) ≡ (p ∧ (p ≡ q)) ≡ p ≡ q ≡ (r ∧ (p ≡ q)) ≡ (p ∧ (p ≡ q)) ≡ p ≡ q
	 =   <Distributivity of ∧ over ≡(3.49), with p,q,r:=r,p,q —replace r∧(p≡q)
		  by r∧p≡r∧q≡r twice>
		r ∧ p ≡ r ∧ q ≡ r ≡ (p ∧ (p ≡ q)) ≡ p ≡ q ≡ r ∧ p ≡ r ∧ q ≡ r ≡ (p ∧ (p ≡ q)) ≡ p ≡ q
	 =   <Distributivity of ∧ over ≡(3.49), with q,r:=p,q —replace p∧(p≡q)
		  by p∧p≡p∧q≡p twice>
		r ∧ p ≡ r ∧ q ≡ r ≡ p ∧ p ≡ p ∧ q ≡ p ≡ p ≡ q ≡ r ∧ p ≡ r ∧ q ≡ r ≡ p ∧ p ≡ p ∧ q ≡ p ≡ p ≡ q
	 =   <Idempotency of ∧(3.38) —replace p∧p by p twice>
		r ∧ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ p ≡ p ≡ q ≡ r ∧ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ p ≡ p ≡ q
	 =   <Symmetry of ≡(3.2), with p,q:=p∧q,p —replace p∧q≡p≡p by p∧q twice>
		r ∧ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ q ≡ r ∧ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ q
	 =   <Golden rule(3.35), with p,q:=r,p —replace r∧p by r≡p≡r∨p twice>
		r ≡ p ≡ r ∨ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ∧ q ≡ r ≡ p ≡ p ∧ q ≡ q
	 =   <Golden rule(3.35), with p,q:=r,q —replace r∧p by r≡q≡r∨q twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ≡ p ∧ q ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ≡ p ∧ q ≡ q
	 =   <Golden rule(3.35) —replace p∧q by p≡q≡p∨q twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ≡ p ≡ q ≡ p ∨ q ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ≡ p ≡ q ≡ p ∨ q ≡ q
	 =   <Symmetry of ≡(3.2), with p,q:=r,p —replace r≡p≡p by r twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ q ≡ p ∨ q ≡ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ q ≡ p ∨ q ≡ q
	 =   <Symmetry of ≡(3.2), with p,q:=p∨q,q —replace p∨q≡q by q≡p∨q twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ q ≡ q ≡ p ∨ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ q ≡ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=r,q —replace r≡q≡q by r twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ∨ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ q ≡ r ∨ q ≡ r ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=q≡r∨q,r —replace q≡r∨q≡r by r≡q≡r∨q twice>
		r ≡ p ≡ r ∨ p ≡ r ≡ r ≡ q ≡ r ∨ q ≡ p ∨ q ≡ r ≡ p ≡ r ∨ p ≡ r ≡ r ≡ q ≡ r ∨ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=r∨p,r —replace r∨p≡r≡r by r∨p twice>
		r ≡ p ≡ r ∨ p ≡ q ≡ r ∨ q ≡ p ∨ q ≡ r ≡ p ≡ r ∨ p ≡ q ≡ r ∨ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=r∨p,q≡r∨q≡p∨q —replace r∨p≡q≡r∨q≡p∨q by q≡r∨q≡p∨q≡r∨p twice>
		r ≡ p ≡ q ≡ r ∨ q ≡ p ∨ q ≡ r ∨ p ≡ r ≡ p ≡ q ≡ r ∨ q ≡ p ∨ q ≡ r ∨ p
	 =   <Symmetry of ≡(3.2), with p,q:=r∨q,p∨q —replace r∨q≡p∨q by p∨q≡r∨q twice>
		r ≡ p ≡ q ≡ p ∨ q ≡ r ∨ q ≡ r ∨ p ≡ r ≡ p ≡ q ≡ p ∨ q ≡ r ∨ q ≡ r ∨ p
	 =   <Symmetry of ≡(3.2), with p,q:=r,q —replace r∨q by q∨r twice>
		r ≡ p ≡ q ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p ≡ r ≡ p ≡ q ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p
	 =   <Symmetry of ≡(3.2), with p,q:=r,p≡q —replace r≡p≡q by p≡q≡r twice>
		p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p ≡ p ≡ q ≡ r ≡ p ∨ q ≡ q ∨ r ≡ r ∨ p
3.39 Prove Definition of ≡(3.52), p ≡ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q).
	 Hint: Apply theorem(3.32), p ∨ q ≡ p ∨ ¬q ≡ p, to the RHS.
	 Answer:
	 (p ∧ q) ∨ (¬p ∧ ¬q)
	 =   <De Morgan(3.47b) —replace ¬p∧¬q by ¬(p∨q)>
		(p ∧ q) ∨ ¬(p ∨ q)
	 =   <(3.32), with p,q:=p∧q,p∨q —replace (p∧q)∨¬(p∨q) by (p∧q)≡(p∧q)∨(p∨q)>
		p ∧ q ≡ (p ∧ q) ∨ (p ∨ q)
	 =   <Distributivity of ∨ over ∨(3.31), with p,q,r:=p∧q,p,q —replace
		  (p∧q)∨(p∨q) by ((p∧q)∨p)∨((p∧q)∨q)>
		p ∧ q ≡ ((p ∧ q) ∨ p) ∨ ((p ∧ q) ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=(p∧q),p —replace (p∧q)∨p by p ∨ (p ∧ q)>
		p ∧ q ≡ (p ∨ (p ∧ q)) ∨ ((p ∧ q) ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=(p∧q),q —replace (p∧q)∨q by q ∨ (p ∧ q)>
		p ∧ q ≡ (p ∨ (p ∧ q)) ∨ (q ∨ (p ∧ q))
	 =   <Absorption(3.43b) —replace p∨(p∧q) by p>
		p ∧ q ≡ p ∨ (q ∨ (p ∧ q))
	 =   <Symmetry of ∧(3.36) —replace p∧q by q∧p>
		p ∧ q ≡ p ∨ (q ∨ (q ∧ p))
	 =   <Absorption(3.43b), with p,q:=q,p —replace q∨(q∧p) by q>
		p ∧ q ≡ p ∨ q
	 =   <Golden rule(3.35) —replace p∧q≡p∨q by p≡q>
		p ≡ q
3.40 Prove Exclusive or(3.53), p ≢ q ≡ (¬p ∧ q) ∨ (p ∧ ¬q). Hint: Try to apply
	 Definition of ≡(3.52).
	 Answer:
	 (¬p ∧ q) ∨ (p ∧ ¬q)
	 =   <Definition of ≡(3.52), with p,q:=¬p,q —replace (¬p∧q)∨(p∧¬q) by ¬p≡q>
		¬p ≡ q
	 =   <Theorem(3.14) —replace ¬p≡q by p≢q>
		p ≢ q
3.41 Prove Implication(3.59), p → q ≡ ¬p ∨ q. At one point of the proof,you may find theorem(3.32) useful.
	 Answer:
	 p → q
	 =   <→(3.57) —replace p→q by p∨q≡q>
		p ∨ q ≡ q
	 =   <Symmetry of ≡(3.2), with p:=p∨q —replace q≡p∨q>
		q ≡ p ∨ q
	 =   <Symmetry of ∨(3.24) —replace p∨q by q∨p>
		q ≡ q ∨ p
	 =   <(3.32), with p,q:=q,p —replace q≡q∨p by q∨¬p>
		q ∨ ¬p
	 =   <Symmetry of ∨(3.24) —replace q∨¬p by ¬p∨q>
		¬p ∨ q
3.42 Prove Implication(3.60), p → q ≡ p ∧ q ≡ p.
	Answer:
	 p → q
	 =   <→(3.57) —replace p→q by p∨q≡q>
		p ∨ q ≡ q
	 =   <Symmetry of ≡(3.2), with p:=p∨q —replace q≡p∨q>
		q ≡ p ∨ q
	 =   <Symmetry of ∨(3.24) —replace p∨q by q∨p>
		q ≡ q ∨ p
	 =   <Golden rule(3.35), with q,p:=p,q —replace q≡q∨p by p≡p∧q>
		p ≡ p ∧ q
	 =   <Symmetry of ≡(3.2) —replace p≡p∧q by p∧q≡p>
		p ∧ q ≡ p
3.43 Prove Contrapositive(3.60), p → q ≡ ¬q → ¬p.
	 Answer:
	 p → q
	 =   <Implication(3.59) —replace p→q by ¬p∨q>
		¬p ∨ q
	 =   <(3.32), with p:=¬p —replace ¬p∨q by ¬p≡¬p∨¬q>
		¬p ≡ ¬p ∨ ¬q
	 =   <Symmetry of ≡(3.2), with p,q:=¬p,¬p∨¬q —replace ¬p≡¬p∨¬q by ¬p∨¬q≡¬p>
		¬p ∨ ¬q ≡ ¬p
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,¬q —replace ¬p∨¬q by ¬q∨¬p>
		¬q ∨ ¬p ≡ ¬p
	 =   <→(3.57), with p,q:=¬q,¬p —replace ¬q∨¬p≡¬p by ¬q→¬p>
		¬q → ¬p
3.44 Prove p → q ≡ ¬(p∧¬q). Axiom(3.57) may not be the best choice to eliminate the
	 implication.
	 Answer:
	 p → q
	 =   <Implication(3.59) —replace p→q by ¬p∨q>
		¬p ∨ q
	 =   <Double negation(3.12), with p:=q —replace q by ¬¬q>
		¬p ∨ ¬¬q
	 =   <De Morgan(3.47a) with q:=¬q —replace ¬p∨¬¬q by ¬(p∧¬q)>
		¬(p ∧ ¬q)
3.45 Prove p → q ≡ ¬p ∨ ¬q ≡ ¬p
	 Answer:
	 p → q
	 =   <Implication(3.59) —replace p→q by ¬p∨q>
		¬p ∨ q
	 =   <(3.32), with p,q:=¬p,q —replace ¬p∨q by ¬p≡¬p∨¬q>
		¬p ≡ ¬p ∨ ¬q
	 =   <Symmetry of ≡(3.2), with p,q:=¬p,¬p∨¬q —replace ¬p≡¬p∨¬q by ¬p∨¬q≡¬p>
		¬p ∨ ¬q ≡ ¬p
3.46 Prove p → q ≡ ¬p ∧ ¬q ≡ ¬q
	 Answer:
	 p → q
	 =   <Implication(3.59) —replace p→q by ¬p∨q>
		¬p ∨ q
	 =   <Symmetry of ∨(3.24), with p:=¬p —replace ¬p∨q by q∨¬p>
		q ∨ ¬p
	 =   <(3.32), with p,q:=q,p —replace q∨¬p by p∨q≡q>
		p ∨ q ≡ q
	 =   <Double negation(3.12), with p:=q —replace q by ¬¬q>
		p ∨ q ≡ ¬¬q
	 =   <Symmetry of ≡(3.2), with p,q:=¬¬q,p∨q —replace p∨q≡¬¬q by ¬¬q≡p∨q>
		¬¬q ≡ p ∨ q
	 =   <Distributivity of ¬ over ≡(3.9), with p,q:=¬q,p∨q —replace ¬¬q≡p∨q
		  by ¬(¬q≡p∨q)>
		¬(¬q ≡ p ∨ q)
	 =   <Symmetry of ≡(3.2), with p,q:=¬q,p∨q —replace ¬q≡p∨q by p∨q≡¬q>
		¬(p ∨ q ≡ ¬q)
	 =   <Distributivity of ¬ over ≡(3.9), with p,q:=p∨q,¬q —replace ¬(p∨q≡¬q)
		  by ¬(p ∨ q) ≡ ¬q>
		¬(p ∨ q) ≡ ¬q
	 =   <De Morgan(3.47b) —replace ¬(p∨q) by ¬p∧¬q>
		¬p ∧ ¬q ≡ ¬q
3.47 Prove Distributivity of → over ≡(3.63), p → (q ≡ r) ≡ p → q ≡ p → r.
	 Answer:
	 p → (q ≡ r)
	 =   <Implication(3.57), with q:=(q≡r) —replace p→(q≡r) by p∨(q≡r)≡q≡r>
		p ∨ (q ≡ r) ≡ q ≡ r
	 =   <Distributivity of ∨ over ≡(3.27) —replace p∨(q≡r) by p∨q≡p∨r>
		p ∨ q ≡ p ∨ r ≡ q ≡ r
	 =   <Symmetry of ≡(3.2), with p:=p∨r —replace p∨r≡q by q≡p∨r>
		p ∨ q ≡ q ≡ p ∨ r ≡ r
	 =   <Implication(3.57) —replace p∨q≡q by p→q>
		p → q ≡ p ∨ r ≡ r
	 =   <Implication(3.57) —replace p∨r≡r by p→r>
		p → q ≡ p → r
3.48 Prove theorem(3.64), p → (q → r) ≡ (p → q) → (p → r).
	 p → (q → r)
	 Answer:
	 =   <Implication(3.59), with q:=q→r —replace p→(q→r) by ¬p∨(q→r)>
		¬p ∨ (q → r)
	 =   <Implication(3.59), with p,q:=q,r —replace q→r by ¬q∨r>
		¬p ∨ (¬q ∨ r)
	 =   <Associativity of ∨(3.25), with p,q:=¬p,¬q —replace ¬p∨(¬q∨r)
		  by (¬p∨¬q)∨r>
		(¬p ∨ ¬q) ∨ r
	 =   <Absorption(3.44b), with p,q:=¬p,¬q —replace (¬p∨¬q) by ¬p∨(¬¬p∧¬q)>
		¬p ∨ (¬¬p ∧ ¬q) ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,(¬¬p∧¬q) —replace ¬p∨(¬¬p∧¬q) by (¬¬p∧¬q)∨¬p>
		(¬¬p ∧ ¬q) ∨ ¬p ∨ r
	 =   <De Morgan(3.47b), with p:=¬p —replace ¬¬p∧¬q by ¬(¬p∨q)>
		¬(¬p ∨ q) ∨ ¬p ∨ r
	 =   <Implication(3.59) —replace ¬p∨q by p→q>
		¬(p → q) ∨ (¬p ∨ r)
	 =   <Implication(3.59) —replace ¬p∨r by p→r>
		¬(p → q) ∨ (p → r)
	 =   <Implication(3.59), with p,q:=p→q,p→r —replace ¬(p→q)∨(p→r)
		  by (p→q)→(p→r)>
		(p → q) → (p → r)
3.49 Prove Currying(3.65), p ∧ q → r ≡ p → (q → r). Use the heuristic of Definition
	 Elimination, (3.23), on page 48. Use of one of (3.59)—(3.61) instead of (3.57)
	 to remove the implication will be more fruitful.
	 Answer:
	 p ∧ q → r
	 =   <Def. of →(3.59), with p,q:=p∧q,r —replace p∧q→r
		  by ¬(p∧q)∨r>
		¬(p ∧ q) ∨ r
	 =   <De Morgan(3.47a) —replace ¬(p∧q) by (¬p ∨ ¬q)>
		(¬p ∨ ¬q) ∨ r
	 =   <Associativity of ∨(3.25), with p,q:=¬p,¬q —replace (¬p∨¬q)∨r
		  by ¬p ∨ (¬q ∨ r)>
		¬p ∨ (¬q ∨ r)
	 =   <Definition of Implication(3.59), with p,q:=q,r —replace ¬q∨r by q→r>
		¬p ∨ (q → r)
	=   <Definition of Implication(3.59), with q:=(q→r) —replace ¬p∨(q→r)
		  by p→(q→r)>
		p → (q → r)
3.50 Prove theorem(3.66), p ∧ (p → q) ≡ p ∧ q. Hint: Try to eliminate the
	 implication in a manner that allows an Absorption law to be used.
	 Answer:
	 p ∧ (p → q)
	 =   <Def. of →(3.59) —replace p→q ¬p∨q>
		p ∧ (¬p ∨ q)
	 =   <Absorption(3.44a) —replace p∧(¬p∨q) by p∧q>
		p ∧ q
3.51 Prove theorem(3.67), p ∧ (q → p) ≡ p. Hint: Try to eliminate the implication
	 in a manner that allows an Absorption law to be used.
	 Answer:
	 p ∧ (q → p)
	 =   <Contrapositive(3.61), with p,q:=q,p —replace ¬p→¬q>
		p ∧ (¬p → ¬q)
	 =   <→(3.59), with p,q:=¬p,¬q —replace ¬p→¬q by ¬¬p∨¬q>
		p ∧ (¬¬p ∨ ¬q)
	 =   <Double negation(3.12) —replace ¬¬p by p>
		p ∧ (p ∨ ¬q)
	 =   <Absorption(3.43a), with q:=¬q —replace p∧(p∨¬q) by p>
		p
3.52 Prove theorem (3.68), p ∨ (p → q) ≡ true. Hint: Use (3.59) to eliminate the
	 implication.
	 Answer:
	 p ∨ (p → q)
	 =   <Def. of →(3.59) —replace p→q by ¬p∨q>
		p ∨ (¬p ∨ q)
	 =   <Associativity of ∨(3.25), with q,r:=¬p,q —replace p∨(¬p∨q) by (p∨¬p)∨q>
		(p ∨ ¬p) ∨ q
	 =   <Metatheorem(3.7), Excluded middle(3.28) and true(3.4) —replace p∨¬p by true>
		true ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=true,q —replace true∨q by q∨true>
		q ∨ true
	 =   <Zero of ∨(3.29), with p:=q —replace q∨true by true>
		true
3.53 Prove theorem(3.69), p ∨ (q → p) ≡ q → p. Hint: use (3.59) to eliminate the
	 implication.
	 Answer:
	 p ∨ (q → p)
	 =   <→(3.59), with p,q:=q,p —replace q→p by ¬q∨p>
		p ∨ (¬q ∨ p)
	 =   <Symmetry of ∨(3.24), with p,q:=p,(¬q∨p) —replace p∨(¬q∨p) by (¬q∨p)∨p>
		(¬q ∨ p) ∨ p
	 =   <Associativity of ∨(3.25), with p,q,r:=¬q,p,p —replace (¬q∨p)∨p
		  by ¬q∨(p∨p)>
		¬q ∨ (p ∨ p)
	 =   <Idempotency of ∨(3.26) —replace p∨p by p>
		¬q ∨ p
	 =   <→(3.59), with p,q:=q,p —replace ¬q∨p by q→p>
		q → p
3.54 Prove theorem(3.70), p ∨ q → p ∧ q ≡ p ≡ q. Hint: Start with p ∨ q → p ∧ q
	 and remove the implication. Head toward a use of the alternative definition
	 of ≡.
	 Answer:
	 p ∨ q → p ∧ q
	 =   <Def. of →(3.59), with p,q:=p∨q,p∧q —replace p∨q→p∧q
		  by ¬(p∨q)∨(p∧q)>
		¬(p ∨ q) ∨ (p ∧ q)
	 =   Symmetry of ∨(3.24), with p,q:=¬(p∨q),(p∧q) —replace ¬(p∨q)∨(p∧q)
		  by (p∧q)∨¬(p∨q)>
		(p ∧ q) ∨ ¬(p ∨ q)
	 =   <De Morgan(3.47b) —replace ¬(p∨q) by (¬p∧¬q)>
		(p ∧ q) ∨ (¬p ∧ ¬q)
	 =   <Def. of ≡(3.52) —replace (p∧q)∨(¬p∧¬q) with p≡q>
		p ≡ q
3.55 Prove Reflexivity of →(3.71), p → p ≡ true
	 Answer:
	 p → p
	 =   <Def. of →(3.60), with q:=p —replace p→p by p∧p≡p>
		p ∧ p ≡ p
	 =   <Idempotency of ∧(3.38) —replace p∧p by p>
		p ≡ p
	 =   <Metatheorem(3.7), Reflexivity of ≡(3.5) and true(3.4) —replace p≡p by true>
		true
3.56 Prove Right zero of →(3.72), p → true ≡ true.
	 Answer:
	 p → true
	 =   <Def. of →(3.59), with q:=true —replace p→true by ¬p∨true>
		¬p ∨ true
	 =   <Zero of ∨(3.29), with p:=¬p —replace ¬p∨true by true>
		true
3.57 Prove left identity of →(3.73), true → p ≡ p.
	 Answer:
	 true → p
	 =   <Def. of →(3.59), with p,q:=true,p —replace true→p by ¬true∨p>
		¬true ∨ p
	 =   <Def. of false(3.8) —replace ¬true by false>
		false ∨ p
	 =   <Symmetry of ∨(3.24), with p,q:=false,p —replace false∨p by p∨false>
		p ∨ false
	 =   <Identity of ∨(3.30) —replace p∨false by p>
		p
3.58 Prove theorem(3.74), p → false ≡ ¬p.
	 Answer:
	 p → false
	 =   <Def. of →(3.59), with p,q:=p,false —replace p→false by ¬p∨false>
		¬p ∨ false
	 =   <Identity of ∨(3.30), with p:=¬p —replace ¬p∨false by ¬p>
		¬p
3.59 Prove theorem(3.75), false → p ≡ true.
	 Answer:
	 =   <Def. of →(3.59), with p,q:=false,p —replace false→p by ¬false∨p>
		¬false ∨ p
	 =   <Negation of false(3.13) —replace ¬false by true>
		true ∨ p
	 =   <Symmetry of ∨(3.24), with p,q:=true,p —replace true∨p by p∨true>
		p ∨ true
	 =   <Zero of ∨(3.29) —replace p∨true by true>
		true
3.60 Prove Weakening/strengthing(3.76a), p → p ∨ q. After eliminating the implication
	 (in a suitable manner), you may find it helpful to use a law of absorption.
	 Answer:
	 p → p ∨ q
	 =   <Def. of →(3.60), with q:=p∨q —replace p→p∨q by p∧(p∨q)≡p>
		p ∧ (p ∨ q) ≡ p —Absorption(3.43a)
3.61 Prove Weakening/strengthening(3.76b), p ∧ q → p. The hint of the preceding
	 exercise applies here also.
	 Answer:
	 p ∧ q → p
	 =   <Def. of →(3.57), with p:=p∧q —replace p∧q→p by (p∧q)∨p≡p>
		(p ∧ q) ∨ p ≡ p
	 =   <Symmetry of ∨(3.24), with p,q:=(p∧q),p —replace (p∧q)∨p by p∨(p∧q)>
		p ∨ (p ∧ q) ≡ p —Absorption(3.43b)
3.62 Prove Weakening/strengthening (3.76c), p ∧ q → p ∨ q. The hint of the preceding
	 exercise applies here also.
	 Answer:
	 p ∧ q → p ∨ q
	 =   <Def. of →(3.59), with p,q:=p∧q,p∨q —replace p∧q→p∨q by ¬(p∧q)∨p∨q>
		¬(p ∧ q) ∨ p ∨ q
	 =   <De Morgan(3.47a) —replace ¬(p∧q) by ¬p ∨ ¬q>
		¬p ∨ ¬q ∨ p ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=¬p∨¬q,p∨q —replace ¬p∨¬q∨p∨q by p∨q∨¬p∨¬q>
		p ∨ q ∨ ¬p ∨ ¬q
	 =   <Symmetry of ∨(3.24), with p,q:=q,¬p —replace q∨¬p by ¬p∨q>
		p ∨ ¬p ∨ q ∨ ¬q
	 =   <Metatheorem(3.7), Excluded middle(3.28) and true(3.4) —replace p∨¬p by true>
		true ∨ q ∨ ¬q
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=q and true(3.4) 
		  —replace q∨¬q by true>
		true ∨ true
	 =   <Zero of ∨(3.29), with p:=true —replace true∨true by true>
		true —theorem(3.4)
3.63 Prove Weakening/strengthening(3.76d), p ∨ (q ∧ r) → p ∨ q. Since the main
	 operator in this expression is ∨, one idea is to remove the implication
	 using (3.57). Alternatively, it can be removed in one step.
	 Answer:
	 p ∨ (q ∧ r) → p ∨ q
	 =   <Def. of →(3.57), with p,q:=p∨(q∧r),p∨q>
		(p ∨ (q ∧ r)) ∨ (p ∨ q) ≡ (p ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=(p∨(q∧r)),(p∨q)>
		(p ∨ q) ∨ (p ∨ (q ∧ r)) ≡ (p ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=(p∨q),p,(q∧r)>
		((p ∨ q) ∨ p) ∨ (q ∧ r) ≡ (p ∨ q)
	 =   <Symmetry of ∨(3.24)>
		((q ∨ p) ∨ p) ∨ (q ∧ r) ≡ (p ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=q,p,p>
		(q ∨ (p ∨ p)) ∨ (q ∧ r) ≡ (p ∨ q)
	 =   <Idempotency of ∨(3.26)>
		(q ∨ p) ∨ (q ∧ r) ≡ (p ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=q,p>
		(p ∨ q) ∨ (q ∧ r) ≡ (p ∨ q)
	 =   <(3.32), with p,q:=(p∨q),q∧r>
		(p ∨ q) ∨ ¬(q ∧ r)
	 =   <De Morgan(3.47a), with p,q:=q,r>
		(p ∨ q) ∨ (¬q ∨ ¬r)
	 =   <Associativity of ∨(3.25), with r:=(¬q∨¬r)>
		p ∨ (q ∨ (¬q ∨ ¬r))
	 =   <Associativity of ∨(3.25), with p,q,r:=q,¬q,¬r>
		p ∨ ((q ∨ ¬q) ∨ ¬r)
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=q and true(3.4)>
		p ∨ (true ∨ ¬r)
	 =   <Symmetry of ∨(3.24), with p,q:=true,¬r>
		p ∨ (¬r ∨ true)
	 =   <Zero of ∨(3.29), with p:=¬r>
		p ∨ true
	 =   <Zero of ∨(3.29)>
		true —theorem(3.4)
3.64 Prove Weakening/strengthening(3.76e), p ∧ q → p ∧ (q ∨ r). Since the main operator
	 in this expression is ∧, one idea is to remove the implication using (3.60).
	 Alternatively, it can be proved in one step.
	 Answer: 
	 p ∧ q → p ∧ (q ∨ r)
	 =   <Def. of →(3.59), with p,q:=p∧q,p∧(q∨r) —replace p∧q→p∧(q∨r)
		  by ¬(p∧q)∨(p∧(q∨r))>
		¬(p ∧ q) ∨ (p ∧ (q ∨ r))
	 =   <De Morgan(3.47a) —replace ¬(p∧q) by (¬p∨¬q)>
		(¬p ∨ ¬q) ∨ (p ∧ (q ∨ r))
	 =   <Distributivity of ∨ over ∧(3.45), with p,q,r:=(¬p∨¬q),p,(q∨r)>
		((¬p ∨ ¬q) ∨ p) ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,¬q —replace ¬p∨¬q by ¬q∨¬p>
		((¬q ∨ ¬p) ∨ p) ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Associativity of ∨(3.25), with p,q,r:=¬q,¬p,p —replace (¬q∨¬p)∨p by ¬q∨(¬p∨p)>
		(¬q ∨ (¬p ∨ p)) ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,p —replace ¬p∨p by p∨¬p>
		(¬q ∨ (p ∨ ¬p)) ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Metatheorem(3.7), Excluded middle(3.28) and true(3.4) —replace p∨¬p by true>
		(¬q ∨ true) ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Zero of ∨(3.29), with p:=¬q —replace ¬q∨true by true>
		true ∧ ((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Symmetry of ∧(3.36), with p,q:=true,((¬p∨¬q)∨(q∨r)), with p,q:=true,((¬p∨¬q)∨(q∨r))
		  —replace true∧((¬p∨¬q)∨(q∨r)) by ((¬p∨¬q)∨(q∨r))true>
		((¬p ∨ ¬q) ∨ (q ∨ r)) ∧ true
	 =   <Identity of ∧(3.39), with p:=((¬p∨¬q)∨(q∨r)) —replace ((¬p∨¬q)∨(q∨r))∧true
		  by ((¬p∨¬q)∨(q∨r))>
		((¬p ∨ ¬q) ∨ (q ∨ r))
	 =   <Associativity of ∨(3.25), with p,q,r:=(¬p∨¬q),q,r —replace ((¬p∨¬q)∨(q∨r))
		  by ((¬p∨¬q)∨q)∨r>
		((¬p ∨ ¬q) ∨ q) ∨ r
	 =   <Associativity of ∨(3.25), with p,q,r:=¬p,¬q,q —replace (¬p∨¬q)∨q
		  by ¬p ∨ (¬q ∨ q)>
		(¬p ∨ (¬q ∨ q)) ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=¬q,q —replace ¬q∨q by q∨¬q>
		(¬p ∨ (q ∨ ¬q)) ∨ r
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=q and true(3.4) —replace q∨¬q by true>
		(¬p ∨ true) ∨ r
	 =   <Zero of ∨(3.29), with p:=¬p —replace ¬p∨true by true>
		true ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=true,r —replace true∨r by r∨true>
		r ∨ true
	 =   <Zero of ∨(3.29), with p:=r —replace r∨true by true>
		true —theorem(3.4)
3.65 Prove Modus ponens, (3.77), p ∧ (p → q) → q. Hint: Use theorem(3.66).
	 p ∧ (p → q) → q
	 =   <(3.66) —replace p∧(p→q) by p∧q>
		p ∧ q → q
	 =   <Def. of →(3.59), with p:=p∧q —replace q → p ∧ q>
		¬(p ∧ q) ∨ q
	 =   <De Morgan(3.47a), —replace ¬(p∧q) by ¬p∨¬q>
		¬p ∨ ¬q ∨ q
	 =   <Symmetry of ∨(3.24), with p:=¬q —replace ¬q∨q by q∨¬q>
		¬p ∨ q ∨ ¬q
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=q and true(3.4)
		  —replace q∨¬q by true>
		¬p ∨ true
	 =   <Zero of ∨(3.29), with p:=¬p —replace ¬p∨true by true>
		true
3.66 Prove theorem(3.78), (p → r) ∧ (q → r) ≡ (p ∨ q → r).
	 (p → r) ∧ (q → r)
	 =   <Def. of →(3.59), with q:=r —replace p→r by ¬p∨r>
		(¬p ∨ r) ∧ (q → r)
	 =   <Def. of →(3.59), with p,q:=q,r —replace q→r by ¬q∨r>
		(¬p ∨ r) ∧ (¬q ∨ r)
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,r —replace ¬p∨r by r∨¬p>
		(r ∨ ¬p) ∧ (¬q ∨ r)
	 =   <Symmetry of ∨(3.24), with p,q:=¬q,r —replace ¬q∨r by r∨¬q>
		(r ∨ ¬p) ∧ (r ∨ ¬q)
	 =   <Distributivity of ∨ over ∧(3.45), with p,q,r:=r,¬p,¬q —replace (r∨¬p)∧(r∨¬q)
		  by r∨(¬p∧¬q)>
		r ∨ (¬p ∧ ¬q)
	 =   <De Morgan(3.47b) —replace (¬p∧¬q) by ¬(p∨q)>
		r ∨ ¬(p ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=r,¬(p∨q) —replace r∨¬(p∨q) by ¬(p∨q)∨r>
		¬(p ∨ q) ∨ r
	 =   <Def. of →(3.59), with p,q:=p∨q,r —replace ¬(p∨q)∨r by p ∨ q → r>
		p ∨ q → r
3.67 Prove theorem(3.79), (p → r) ∧ (¬p → r) ≡ r.
	 (p → r) ∧ (¬p → r)
	 =   <Def. of →(3.59), with q:=r —replace p→r by ¬p∨r>
		(¬p ∨ r) ∧ (¬p → r)
	 =   <Def. of →(3.59), with p,q:=¬p,r —replace ¬p→r by ¬¬p∨r>
		(¬p ∨ r) ∧ (¬¬p ∨ r)
	 =   <Double negation(3.12) —replace ¬¬p by p>
		(¬p ∨ r) ∧ (p ∨ r)
	 =   <Symmetry of ∨(3.24), with p,q:=¬p,r —replace ¬p∨r by r∨¬p>
		(r ∨ ¬p) ∧ (p ∨ r)
	 =   <Symmetry of ∨(3.24), with p,q:=p,r —replace p∨r by r∨p>
		(r ∨ ¬p) ∧ (r ∨ p)
	 =   <Distributivity of ∨ over ∧(3.45), with p,q,r:=r,¬p,p —replace (r∨¬p)∧(r∨p)
		  by r∨(¬p∧p)>
		r ∨ (¬p ∧ p)
	 =   <Symmetry of ∧(3.36), with p,q:=¬p,p —replace ¬p∧p by p∧¬p>
		r ∨ (p ∧ ¬p)
	 =   <Contradiction(3.42) —replace p∧¬p by false>
		r ∨ false
	 =   <Identity of ∨(3.30), with p:=r —replace r∨false by r>
		r
3.68 Prove mutual implication(3.80). Begin by replacing each conjunct in the LHS
	 using (3.59) and then use the Golden rule.
	 Answer:
	 (p → q) ∧ (q → p)
	 =   <Def. of →(3.59) —replace p→q by ¬p∨q>
		(¬p ∨ q) ∧ (q → p)
	 =   <Def. of →(3.59), with p,q:=q,p —replace q→p by ¬q∨p>
		(¬p ∨ q) ∧ (¬q ∨ p)
	 =   <Distributivity of ∧ over ∨(3.46), with p,q,r:=(¬p∨q),¬q,p —replace
		  (¬p∨q)∧(¬q∨p) by ((¬p∨q)∧¬q) ∨ ((¬p∨q)∧p)>
		((¬p ∨ q) ∧ ¬q) ∨ ((¬p ∨ q) ∧ p)
	 =   <Symmetry of ∧(3.36), with p,q:=(¬p∨q),¬q —replace (¬p∨q)∧¬q by ¬q∧(¬p∨q)>
		(¬q ∧ (¬p ∨ q)) ∨ ((¬p ∨ q) ∧ p)
	 =   <Symmetry of ∧(3.36), with p,q:=(¬p∨q),p —replace (¬p∨q)∧p by p∧(¬p∨q)>
		(¬q ∧ (¬p ∨ q)) ∨ (p ∧ (¬p ∨ q))
	 =   <Distributivity of ∧ over ∨(3.46), with p,q,r:=¬q,¬p,q —replace ¬q∧(¬p∨q)
		  by (¬q∧¬p)∨(¬q∧q)>
		((¬q ∧ ¬p) ∨ (¬q ∧ q)) ∨ (p ∧ (¬p ∨ q))
	 =   <Distributivity of ∧ over ∨(3.46), with p,q,r:=p,¬p,q —replace p∧(¬p∨q)
		  by (p∧¬p)∨(p∧q)>
		((¬q ∧ ¬p) ∨ (¬q ∧ q)) ∨ ((p ∧ ¬p) ∨ (p ∧ q))
	 =   <Symmetry of ∧(3.36), with p,q:=¬q,q —replace ¬q∧q by q∧¬q>
		((¬q ∧ ¬p) ∨ (q ∧ ¬q)) ∨ ((p ∧ ¬p) ∨ (p ∧ q))
	 =   <Contradiction(3.42), with p:=q —replace q∧¬q by false>
		((¬q ∧ ¬p) ∨ false) ∨ ((p ∧ ¬p) ∨ (p ∧ q))
	 =   <Contradiction(3.42) —replace p∧¬p by false>
		((¬q ∧ ¬p) ∨ false) ∨ (false ∨ (p ∧ q))
	 =   <Symmetry of ∨(3.24), with p,q:=false,(p∧q) —replace false∨(p∧q)
		  by (p∧q)∨false>
		((¬q ∧ ¬p) ∨ false) ∨ ((p ∧ q) ∨ false)
	 =   <Identity of ∨(3.30), with p:=¬q∧¬p —replace (¬q∧¬p)∨false by ¬q∧¬p>
		(¬q ∧ ¬p) ∨ ((p ∧ q) ∨ false)
	 =   <Identity of ∨(3.30), with p:=p∧q —replace (p∧q)∨false by p∧q>
		(¬q ∧ ¬p) ∨ (p ∧ q)
	 =   <De Morgan(3.47b), with p,q:=q,p —replace ¬q∧¬p by ¬(q∨p)>
		¬(q ∨ p) ∨ (p ∧ q)
	 =   <(3.32), with p,q:=(p∧q),(q∨p) —replace ¬(q∨p)∨(p∧q)
		  by p∧q≡(p∧q)∨(q∨p)>
		p ∧ q ≡ (p ∧ q) ∨ (q ∨ p)
	 =   <Golden rule(3.35) —replace p∧q by p≡q≡p∨q>
		p ∧ q ≡ (p ≡ q ≡ p∨q) ∨ (q ∨ p)
	 =   <Symmetry of ∨(3.24), with p,q:=(p≡q≡p∨q),(q∨p) —replace (p≡q≡p∨q)∨(q∨p)
		  by (q∨p)∨(p≡q≡p∨q)>
		p ∧ q ≡ (q ∨ p) ∨ (p ≡ q ≡ p∨q)
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=(q∨p),p≡q,p∨q —replace
		 (q∨p)∨(p≡q≡p∨q) by ((q∨p)∨(p≡q))≡((q∨p)∨(p∨q))>
		p ∧ q ≡ ((q ∨ p) ∨ (p ≡ q)) ≡ ((q ∨ p) ∨ (p ∨ q))
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=q∨p,p,q —replace (q∨p)∨(p≡q)
		  by ((q∨p)∨p)≡((q∨p)∨q)>
		p ∧ q ≡ ((q ∨ p) ∨ p) ≡ ((q ∨ p) ∨ q) ≡ ((q ∨ p) ∨ (p ∨ q))
	 =   <Distributivity of ∨ over ∨(3.31), with p,q,r:=q∨p,p,q —replace ((q∨p)∨(p∨q))
		  by ((q∨p)∨p)∨((q∨p)∨q)>
		p ∧ q ≡ ((q ∨ p) ∨ p) ≡ ((q ∨ p) ∨ q) ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=q,p,p —replace (q∨p)∨p by q∨(p∨p)>
		p ∧ q ≡ (q ∨ (p ∨ p)) ≡ ((q ∨ p) ∨ q) ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=q,p —replace q∨p by p∨q>
		p ∧ q ≡ (q ∨ (p ∨ p)) ≡ ((p ∨ q) ∨ q) ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=p,q,q —replace (p∨q)∨q by p∨(q∨q)>
		p ∧ q ≡ (q ∨ (p ∨ p)) ≡ (p ∨ (q ∨ q)) ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Idempotency of ∨(3.26) —replace p∨p by p>
		p ∧ q ≡ q ∨ p ≡ (p ∨ (q ∨ q)) ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Idempotency of ∨(3.26), with p:=q —replace q∨q by q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ ((q ∨ p) ∨ p) ∨ ((q ∨ p) ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=q,p,p —replace (q∨p)∨p by q∨(p∨p)>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ (p ∨ p)) ∨ ((q ∨ p) ∨ q)
	 =   <Symmetry of ∨(3.24), with p,q:=q,p —replace q∨p by p∨q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ (p ∨ p)) ∨ ((p ∨ q) ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=p,q,q —replace (p∨q)∨q by p∨(q∨q)>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ (p ∨ p)) ∨ (p ∨ (q ∨ q))
	 =   <Idempotency of ∨(3.26) —replace p∨p by p>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ p) ∨ (p ∨ (q ∨ q))
	 =   <Idempotency of ∨(3.26), with p:=q —replace q∨q by q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ p) ∨ (p ∨ q)
	 =   <Associativity of ∨(3.25), with p,q,r:=q∨p,p,q —replace (q∨p)∨(p∨q)
		  by ((q ∨ p) ∨ p) ∨ q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ ((q ∨ p) ∨ p) ∨ q
	 =   <Associativity of ∨(3.25), with p,q,r:=q,p,p —replace (q∨p)∨p by q∨(p∨p)>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ (p ∨ p)) ∨ q
	 =   <Idempotency of ∨(3.26) —replace p∨p by p>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (q ∨ p) ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=q,p —replace q∨p by p∨q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ (p ∨ q) ∨ q
	 =   <Associativity of ∨(3.25), with p,q,r:=p,q,q —replace (p∨q)∨q by p∨(q∨q)>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ p ∨ (q ∨ q)
	 =   <Idempotency of ∨(3.26),with p:=q —replace q∨q by q>
		p ∧ q ≡ q ∨ p ≡ p ∨ q ≡ p ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=q,p —replace q∨p by p∨q>
		p ∧ q ≡ p ∨ q ≡ p ∨ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=(p∧q),(p∨q) —replace p∧q≡p∨q≡p∨q
		  by p∧q>
		p ∧ q ≡ p ∨ q
	 =   <Golden rule(3.35) —replace p∧q by p ≡ q ≡ p ∨ q>
		p ≡ q ≡ p ∨ q ≡ p ∨ q
	 =   <Symmetry of ≡(3.2), with p,q:=q,p ∨ q —replace q≡p∨q≡p∨q by q>
		p ≡ q
3.69 Prove Antisymmetry(3.81) in two steps (use Mutual implication(3.80)).
	 Answer:
	 (p → q) ∧ (q → p) → (p ≡ q)
	 =   <Mutual implication(3.80) —replace (p→q)∧(q→p) by (p≡q)>
		(p ≡ q) → (p ≡ q)
	 =   <Reflexivity of →(3.71),with p:=p≡q —replace (p≡q)→(p≡q)
		  by true>
		true —theorem(3.4)
3.70 Prove Transitivity of implication(3.82a). Start with the whole expression,
	 transform each of the four implications in it using (3.59), and then massage.
	 Answer:
	 (p → q) ∧ (q → r) → (p → r)
	 =   <Def. of →(3.59), with p,q:=(p→q)∧(q→r),(p→r) —replace (p→q)∧(q→r)→(p→r)
		  by ¬((p→q)∧(q→r))∨(p→r)>
		¬((p → q) ∧ (q → r)) ∨ (p → r)
	 =   <Def. of →(3.59) —replace p→q by ¬p∨q>
		¬((¬p ∨ q) ∧ (q → r)) ∨ (p → r)
	 =   <Def. of →(3.59), with p,q:=q,r —replace q→r by ¬q∨r>
		¬((¬p ∨ q) ∧ (¬q ∨ r)) ∨ (p → r)
	 =   <Def. of →(3.59), with q:=r —replace p→r by ¬p∨r>
		¬((¬p ∨ q) ∧ (¬q ∨ r)) ∨ ¬p ∨ r
	 =   <De Morgan(3.47a), with p,q:=(¬p∨q),(¬q∨r) —replace ¬((¬p∨q)∧(¬q∨r))
		  by ¬(¬p∨q)∨¬(¬q∨r)>
		¬(¬p ∨ q) ∨ ¬(¬q ∨ r) ∨ ¬p ∨ r
	 =   <De Morgan(3.47b), with p:=¬p —replace ¬(¬p∨q) by (¬¬p∧¬q)>
		(¬¬p ∧ ¬q) ∨ ¬(¬q ∨ r) ∨ ¬p ∨ r
	 =   <De Morgan(3.47b), with p,q:=¬q,r —replace ¬(¬q∨r) by (¬¬q∧¬r)>
		(¬¬p ∧ ¬q) ∨ (¬¬q ∧ ¬r) ∨ ¬p ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=(¬¬p∧¬q)∨(¬¬q∧¬r),¬p —replace
		  (¬¬p∧¬q)∨(¬¬q∧¬r)∨¬p by ¬p∨(¬¬p∧¬q)∨(¬¬q∧¬r)>
		¬p ∨ (¬¬p ∧ ¬q) ∨ (¬¬q ∧ ¬r) ∨ r
	 =   <Absorption(3.44b), with p,q:=¬p,¬q —replace ¬p∨(¬¬p∧¬q) by ¬p∨¬q>
		¬p ∨ ¬q ∨ (¬¬q ∧ ¬r) ∨ r
	 =   <Absorption(3.44b), with p,q:=¬q,¬r —replace ¬q∨(¬¬q∧¬r) by ¬q∨¬r>
		¬p ∨ r ∨ ¬q ∨ ¬r
	 =   <Symmetry of ∨(3.24), with p,q:=r,¬q —replace r∨¬q by ¬q∨r>
		¬p ∨ ¬q ∨ r ∨ ¬r
	 =   <Metatheorem(3.7), Excluded middle(3.28), with p:=r and true(3.4)
		  —replace r∨¬r by true>
		¬p ∨ ¬q ∨ true
	 =   <Zero of ∨(3.29), with p:=¬p∨¬q —replace ¬p∨¬q∨true by true>
		true —theorem(3.4)
3.71 Prove Transitivity(3.82b), (p ≡ q) ∧ (q → r) → (p → r). Use Mutual
	 implication(3.80), Transitivity(3.82a), and Currying(3.65).
	 Answer:
	 =   <Mutual implication(3.80) —replace p≡q by (p→q)∧(q→p)>
		(p → q) ∧ (q → p) ∧ (q → r) → (p → r)
	 =   <Symmetry of ∧(3.36), with p,q:=(p→q),(q→p) —replace (p→q)∧(q→p)
		  by (q→p)∧(p→q)>
		(q → p) ∧ (p → q) ∧ (q → r) → (p → r)
	 =   <Currying(3.65), with p,q,r:=(q→p),(p→q)∧(q→r),(p→r) —replace
		  (q→p)∧(p→q)∧(q→r)→(p→r) by (q→p)→((p→q)∧(q→r)→(p→r))>
		(q → p) → ((p → q) ∧ (q → r) → (p → r))
	 =   <Metatheorem(3.7), Transitivity(3.82a) and True(3.4) —replace
		  (p→q)∧(q→r)→(p→r) by true>
		(q → p) → true
	 =   <Right zero of →(3.72), with p:=(q→p) —replace (q→p)→true by true>
		true —theorem(3.4)
3.72 Prove Transitivity(3.82c), (p → q) ∧ (q ≡ r) → (p → r). Use Mutual
	 implication(3.80), Transitivity(3.82a), and Currying(3.65).
	 (p → q) ∧ (q ≡ r) → (p → r)
	 Answer:
	 =   <Mutual implication(3.80), with p,q:=q,r —replace q≡r by (q→r)∧(r→q)>
		(p → q) ∧ (q → r) ∧ (r → q) → (p → r)
	 =   <Symmetry of ∧(3.36), with p,q:=(p→q)∧(q→r),(r→q) —replace (p→q)∧(q→r)∧(r→q)
		  by (r→q)∧(p→q)∧(q→r)>
		(r → q) ∧ (p → q) ∧ (q → r) → (p → r)
	 =   <Currying(3.65), with p,q,r:=(r→q),(p→q)∧(q→r),(p→r) —replace
		  (r→q)∧(p→q)∧(q→r)→(p→r) by (r→q)→((p→q)∧(q→r)→(p→r))>
		(r → q) → ((p → q) ∧ (q → r) → (p → r))
	 =   <Metatheorem(3.7), Transitivity(3.82a) and True(3.4) —replace
		  (p→q)∧(q→r)→(p→r) by true>
		(r → q) → true
	 =   <Right zero of →(3.72), with p:=(r→q) —replace (r→q)→true by true>
		true —theorem(3.4)
3.73 Prove Substitution(3.84a), (e = f) ∧ E[z:=e] ≡ (e = f) ∧ E[z:=f]. Begin with Leibniz(3.83)
	 and replace the implication.
	 Answer:
	 (e = f) → (E[z:=e]=E[z:=f])
	 =   <(3.62), with p,q,r:=(e=f),E[z:=e],E[z:=f] —replace (e=f)→(E[z:=e]=E[z:=f])
		by (e=f)∧E[z:=e]≡(e=f)∧E[z:=f]>
		(e = f) ∧ E[z:=e] ≡ (e = f) ∧ E[z:=f]
3.74 Prove Substitution(3.84b), (e = f) → E[z:=e] ≡ (e = f) → E[z:=f].
	 Answer:
	 (e = f) → (E[z:=e] = E[z:=f])
	 =   <Distributivity of → over ≡(3.63), with p,q,r:=(e=f),E[z:=e],E[z:=f] —replace
		  (e=f)→(E[z:=e]=E[z:=f]) by (e=f)→E[z:=e]≡(e=f)→E[z:=f]>
		(e = f) → E[z:=e] ≡ (e = f) → E[z:=f]
3.75 Prove Substitution(3.84c), q ∧ (e = f) → E[z:=e] ≡ q ∧ (e = f) → E[z:=f]. Using
	 Currying(3.65).
	 Answer:
	 q ∧ (e = f) → E[z:=e] ≡ q ∧ (e = f) → E[z:=f]
	 =   <Currying(3.65), with p,q,r:=q,(e=f),E[z:=e] —replace q∧(e=f)→E[z:=e]
		  by q→((e=f)→E[z:=e])>
		q → ((e = f) → E[z:=e]) ≡ q ∧ (e = f) → E[z:=f]
	 =   <Currying(3.65), with p,q,r:=q,(e=f),E[z:=f] —replace q∧(e=f)→E[z:=f]
		  by q→((e=f)→E[z:=f])>
		q → ((e = f) → E[z:=e]) ≡ q → ((e = f) → E[z:=f])
	 =   <Substitution(3.84b) —replace (e=f)→E[z:=f] by (e=f)→E[z:=e]>
		q → ((e = f) → E[z:=e]) ≡ q → ((e = f) → E[z:=e])
	 =   <Metatheorem(3.7), Reflexivity of ≡(3.5) and True(3.4) —replace
		q→((e=f)→E[z:=e])≡q→((e=f)→E[z:=e]) by true>
		true
3.76 Prove Replace by true(3.85a), p → E[z:=p] ≡ p → E[z:=true]. In order to be able
	 to use (3.84b), introduce the equivalent true into the antecedent.
	 Answer:
	 p → E[z:=p] ≡ p → E[z:=true]
	 =   <Identity of ≡(3.3), with q:=p —replace p by true≡p twice>
		(true ≡ p) → E[z:=p] ≡ (true ≡ p) → E[z:=true]
	 =   <Symmetry of ≡(3.2), with p,q:=true,p —replace true≡p by p≡true twice>
		(p ≡ true) → E[z:=p] ≡ (p ≡ true) → E[z:=true]
	 =   <Metatheorem(3.7), Substitution(3.84a) and True(3.4) —replace
		(p≡true)→E[z:=p]≡(p≡true)→E[z:=true] by true>
		true
3.77 Prove Replace by true(3.85b), q ∧ p → E[z:=p] ≡ q ∧ p → E[z:=true].
	 Answer:
	 q ∧ p → E[z:=p] ≡ q ∧ p → E[z:=true]
	 =   <Identity of ≡(3.3), with q:=p —replace p by true≡p twice>
		q ∧ (true ≡ p) → E[z:=p] ≡ q ∧ (true ≡ p) → E[z:=true]
	 =   <Metatheorem(3.7), Substitution(3.84c) and True(3.4) —replace
		  q∧(true≡p)→E[z:=p]≡q∧(true≡p)→E[z:=true] by true>
		true
3.78 Prove Replace by false (3.86a), E[z:=p] → p ≡ E[z:=false] → p.
	 Answer:
	 E[z:=p] → p ≡ E[z:=false] → p
	 =   <Contrapositive(3.61), with p,q:=E[z:=p],p —replace E[z:=p]→p
		  by ¬p→¬E[z:=p]>
		¬p → ¬E[z:=p] ≡ E[z:=false] → p
	 =   <Contrapositive(3.61), with p,q:=E[z:=false],p —replace E[z:=false]→p
		  by ¬p→¬E[z:=false]>
		¬p → ¬E[z:=p] ≡ ¬p → ¬E[z:=false]
	 =   <Double negation(3.12) —replace p by ¬¬p>
		¬p → ¬E[z:=¬¬p] ≡ ¬p → ¬E[z:=false]
	 =   <Def. of false(3.8) —replace false by ¬true>
		¬p → ¬E[z:=¬¬p] ≡ ¬p → ¬E[z:=¬true]
	 =   <Let F=¬E[z:=¬z] —replace ¬p→¬E[z:=¬¬p]≡¬p→¬E[z:=¬true]
		  by ¬p→F[z:=¬p]≡¬p→F[z:=true]>
		¬p → F[z:=¬p] ≡ ¬p → F[z:=true]
	 =   <Metatheorem(3.7), Substitution(3.85a), with p,E:=¬p,F and True(3.4) —replace
		  ¬p→F[z:=¬p]≡¬p→F[z:=true] by true>
		true
3.79 Prove Replace by false(3.86b), E[z:=p] → p ∨ q ≡ E[z:=false] → p ∨ q.
	 Answer:
	 E[z:=p] → p ∨ q ≡ E[z:=false] → p ∨ q
	 =   <Contrapositive(3.61), with p,q:=E[z:=p],p∨q —replace E[z:=p]→p∨q by ¬(p∨q)→E[z:=p]>
		¬(p ∨ q) → ¬E[z:=p] ≡ E[z:=false] → p ∨ q
	 =   <Contrapositive(3.61), with p,q:=E[z:=false],p∨q —replace E[z:=false]→p∨q by ¬(p∨q)→¬E[z:=false]>
		¬(p ∨ q) → ¬E[z:=p] ≡ ¬(p ∨ q) → ¬E[z:=false]
	 =   <De Morgan(3.47b) —replace ¬(p∨q) by ¬p∧¬q twice>
		¬p ∧ ¬q → ¬E[z:=p] ≡ ¬p ∧ ¬q → ¬E[z:=false]
	 =   <Symmetry of ∧(3.36), with p,q:=¬p,¬q —replace ¬p∧¬q by ¬q∧¬p twice>
		¬q ∧ ¬p → ¬E[z:=p] ≡ ¬q ∧ ¬p → ¬E[z:=false]
	 =   <Double negation(3.12) —replace p by ¬¬p>
		¬q ∧ ¬p → ¬E[z:=¬¬p] ≡ ¬q ∧ ¬p → ¬E[z:=false]
	 =   <Def. of false(3.8) —replace false by ¬true>
		¬q ∧ ¬p → ¬E[z:=¬¬p] ≡ ¬q ∧ ¬p → ¬E[z:=¬true]
	 =   <Let F=¬E[z:=¬z] —replace ¬q∧¬p→¬E[z:=¬¬p]≡¬q∧¬p→¬E[z:=¬true]
		  by ¬q∧¬p→F[z:=¬p]≡¬q∧¬p→F[z:=true]>
		¬q ∧ ¬p → F[z:=¬p] ≡ ¬q ∧ ¬p → F[z:=true]
	 =   <Metatheorem(3.7), Substitution(3.85b), with p,q,E:=¬q,¬p,F and True(3.4) —replace
		  ¬q∧¬p→F[z:=¬p]≡¬q∧¬p→F[z:=true] by true>
		true
3.80 Prove Replace by true(3.87), p ∧ E[z:=p] ≡ p ∧ E[z:=true]
	 Answer:
	 p ∧ E[z:=p] ≡ p ∧ E[z:=true]
	 =   <Identity of ≡(3.3) —replace p by true≡p twice>
		(true ≡ p) ∧ E[z:=p] ≡ (true ≡ p) ∧ E[z:=true]
	 =   <Symmetry of ≡(3.2), with p,q:=true,p —replace true≡p by p≡true twice>
		(p ≡ true) ∧ E[z:=p] ≡ (p ≡ true) ∧ E[z:=true]
	 =   <Metatheorem(3.7), Substitution(3.85a), with e,f:=p,true and True(3.4) —replace
		  (p≡true)∧E[z:=p]≡(p≡true)∧E[z:=true] by true>
		true
3.81 Prove Replace by false(3.88), p ∨ E[z:=p] ≡ p ∨ E[z:=false].
	 Answer:
	 p ∨ E[z:=p] ≡ p ∨ E[z:=false]
	 =   <Double negation(3.12), with p:=p∨E[z:=p] —replace p∨E[z:=p] by ¬¬(p∨E[z:=p])>
		¬¬(p ∨ E[z:=p]) ≡ p ∨ E[z:=false]
	 =   <Double negation(3.12), with p:=p∨E[z:=false] —replace p∨E[z:=false] by ¬¬(p∨E[z:=false])>
		¬¬(p ∨ E[z:=p]) ≡ ¬¬(p ∨ E[z:=false])
	 =   <De Morgan(3.47b), with p,q:=p,E[z:=p] —replace ¬(p∨E[z:=p]) by (¬p ∧ ¬E[z:=p])>
		¬(¬p ∧ ¬E[z:=p]) ≡ ¬¬(p ∨ E[z:=false])
	 =   <De Morgan(3.47b), with p,q:=p,E[z:=false] —replace ¬(p∨E[z:=false]) by (¬p∧¬E[z:=false])>
		¬(¬p ∧ ¬E[z:=p]) ≡ ¬(¬p ∧ ¬E[z:=false])
	 =   <Distributivity of ¬ over ≡(3.9), with p,q:=¬(¬p∧¬E[z:=p]),¬(¬p∧¬E[z:=false])
		  —replace ¬(¬p∧¬E[z:=p])≡¬(¬p∧¬E[z:=false]) by ¬(¬p ∧ ¬E[z:=p] ≡ ¬(¬p ∧ ¬E[z:=false]))>
		¬(¬p ∧ ¬E[z:=p] ≡ ¬(¬p ∧ ¬E[z:=false]))
	 =   <Symmetry of ≡(3.2), with p,q:=¬p∧¬E[z:=p],¬(¬p∧¬E[z:=false]) —replace
		  ¬p∧¬E[z:=p]≡¬(¬p∧¬E[z:=false]) by ¬(¬p∧¬E[z:=false])≡¬p∧¬E[z:=p]>
		¬(¬(¬p ∧ ¬E[z:=false]) ≡ ¬p ∧ ¬E[z:=p])
	 =   <Distributivity of ¬ over ≡(3.9), with p,q:=¬(¬p∧¬E[z:=false])≡¬p∧¬E[z:=p] —replace
		  ¬(¬p∧¬E[z:=false])≡¬p∧¬E[z:=p] by ¬(¬p∧¬E[z:=false]≡¬p∧¬E[z:=p])>
		¬¬(¬p ∧ ¬E[z:=false] ≡ ¬p ∧ ¬E[z:=p])
	 =   <Double negation(3.12), with p:=¬p∧¬E[z:=false]≡¬p ∧¬E[z:=p] —replace
		  ¬¬(¬p∧¬E[z:=false]≡¬p∧¬E[z:=p]) by ¬p ∧ ¬E[z:=false] ≡ ¬p ∧ ¬E[z:=p]>
		¬p ∧ ¬E[z:=false] ≡ ¬p ∧ ¬E[z:=p]
	 =   <Symmetry of ≡(3.2), with p,q:=¬p∧¬E[z:=false],¬p∧¬E[z:=p] —replace
		  ¬p∧¬E[z:=false]≡¬p∧¬E[z:=p] by ¬p∧¬E[z:=p]≡¬p∧¬E[z:=false]>
		¬p ∧ ¬E[z:=p] ≡ ¬p ∧ ¬E[z:=false]
	 =   <Double negation(3.12) —replace p by ¬¬p>
		¬p ∧ ¬E[z:=¬¬p] ≡ ¬p ∧ ¬E[z:=false]
	 =   <Def. of false(3.8) —replace false by ¬true>
		¬p ∧ ¬E[z:=¬¬p] ≡ ¬p ∧ ¬E[z:=¬true]
	 =   <Let f=¬E[z:=¬z] —replace ¬p∧¬E[z:=¬¬p]≡¬p∧¬E[z:=¬true] by ¬p[F:=z¬p]≡¬p∧F[z:=true]>
		¬p ∧ F[z:=¬p] ≡ ¬p ∧ F[z:=true]
	 =   <Metatheorem(3.7), Replace by true(3.87), with p,E:=¬p,F and True(3.4) —replace
		  ¬p∧F[z:=¬p] ≡ ¬p∧F[z:=true] by true>
		true
3.82 Prove p → (q → p) using theorem (3.85a) in the first step.
	 Answer:
	 p → (q → p)
	 =   <Replace by true(3.85a) —replace q→p by q→true>
		p → (q → true)
	 =   <Right zero of →(3.72), with p:=q —replace q→true by true>
		p → true
	 =   <Right zero of →(3.72) —replace p→true by true>
		true —theorem(3.4)
3.83 Prove Shannon(3.89),E[z:=p] ≡ (p ∧ E[z:=true]) ∨ (¬p ∧ E[z:=false]).
	 Answer:
	 Case analysis:
	 Let R = (E[z:=p] ≡ ((p ∧ E[z:=true]) ∨ (¬p ∧ E[z:=false])))
	 p → R
	 =   <(3.3)>
		(p ≡ true) → (R ≡ true)
	 =   <(3.3)>
		(p ≡ true) → (R ≡ (E[z:=true]≡E[z:=true]))
	 =   <(3.30), (3.39), (3.40)>
		(p ≡ true) → (R ≡ (E[z:=true] ≡ ((true ∧ E[z:=true]) ∨ (¬true ∧ E[z:=false]))))
	 =
		(p ≡ true) → (R ≡ (E[z:=p] ≡ ((p ∧ E[z:=p]) ∨ (¬p ∧ E[z:=false])))[p:=true])
	 =
		(p ≡ true) → (R ≡ R[p:=true])
	 =
		true (Lemma01)
	 ¬p → R
	 =   <(3.3)>
		(p ≡ false) → (R ≡ true)
	 =   <(3.3)>
		(p ≡ false) → (R ≡ (E[z:=false]≡E[z:=false]))
	 =   <(3.30), (3.39), (3.40)>
		(p ≡ false) → (R ≡ (E[z:=false] ≡ ((false ∧ E[z:=false]) ∨ (¬false ∧ E[z:=false]))))
	 =
		(p ≡ false) → (R ≡ (E[z:=p] ≡ ((p ∧ E[z:=p]) ∨ (¬p ∧ E[z:=false])))[p:=false])
	 =
		(p ≡ false) → (R ≡ R[p:=false])
	 =
		true (Lemma02)
	 R
	 =   <(3.79)>
		(p → R) ∧ (¬p → R)
	 =   <Lemma01 —replace (p→R) by true, Lemma02 —replace (¬p→R) by true>
		true ∧ true
	 =   <Identity of ∧(3.39), with p:=true —replace true∧true by true>
		true
3.84 Prove Weakening/strengthening (3.76e), p ∧ q → p ∧ (q ∨ r), using Replace by true(3.85b).
q ∧ p → E[z:=p] ≡ q ∧ p → E[z:=true]
	 Answer:
	 p ∧ q → p ∧ (q ∨ r)
	 =   <Replace by true(3.85b), with p,q:=q,p —replace p∧(q∨r) by true∧(q∨r)>
		p ∧ q → p ∧ (true ∨ r)
	 =   <Symmetry of ∨(3.24), with p,q:=true,r>
		p ∧ q → p ∧ (r ∨ true)
	 =   <Zero of ∨(3.29), with p:=r —replace r∨true by true>
		p ∧ q → p ∧ true
	 =   <Identity of ∧(3.39), —replace p∧true by p>
		p ∧ q → p —Weakening/strengthening(3.76b)
3.85 Consider any expression P of the form true, q ∧ r, q ≡ r, or q → r, and consider
	 its dual PD (see Def. (2.2) on page 31). Prove that P ≡ ¬PD for expressions
	 of the form given above, provided it holds for their subexpressions.
	 Hint: By the definition of the dual, for an operation like ∧, (q ∧ r)D ≡ qD ∨ rD.
	 Answer:
	 Let P=q ∧ r
	 ¬PD
	 =   <Def. of  (2.2) —replace q∧r by qD ∨ rD>
		¬(qD ∨ rD)
	 =   <Inductive assumption (provided it holds for their subexpressions)>
		¬(¬q ∨ ¬r)
	 =   <De Morgan(3.47a), with p,q:=q,r —replace ¬q ∨ ¬r by ¬(q ∧ r)>
		¬¬(q ∧ r)
	 =   <Double negation(3.12), with p:=(q∧r) —replace ¬¬(q∧r) by q∧r>
		P
	Let P=q ∨ r
	 ¬PD
	 =   <Def. of Duality(2.2) —replace q∨r by qD ∧ rD>
		¬(qD ∧ rD)
	 =   <Inductive assumption (provided it holds for their subexpressions)>
		¬(¬q ∧ ¬r)
	 =   <De Morgan(3.47b), with p,q:=q,r —replace ¬q ∧ ¬r by ¬(q ∨ r)>
		¬¬(q ∨ r)
	 =   <Double negation(3.12), with p:=(q∨r) —replace ¬¬(q∨r) by q∨r>
		P
	Let P=q → r
	 ¬PD
	 =   <Def. of Duality(2.2) —replace q→r by q⇍r>
		¬(qD ⇍ rD)
	 =   <Inductive assumption (provided it holds for their subexpressions)>
		¬(¬q ⇐ ¬r)
	 =
		¬q ⇐ ¬r
	 =
		P
	Let P=q ≡ r
	 ¬PD
	 =   <Def. of Duality(2.2) —replace q≡r by q≢r>
		¬(qD ≢ rD)
	 =   <Inductive assumption (provided it holds for their subexpressions)>
		¬(¬q ≡ ¬r)
	 =
		q ≡ r
	 =
		P
3.88
Conjunctive normal form
	E0 ∧ … En-1
	Each Ei is a disjunction of variables and negations of variables.
		(a ∨ ¬b) ∧ (a ∨ b ∨ c) ∧ (¬a)
Disjunctive normal form
	E0 ∨ … En-1
	Each Ei is a conjunction of variables and negations of variables.
		(a ∧ ¬b) ∨ (a ∧ b ∧ c) ∨ (¬a)
false < true
	minterm: v0 ∧ … ∧ vn
	maxterm: v0 ∨ … ∨ vn
∧ conjunction(and) multiplies, minterm
∨ disjunction(inclusive or) maxterm
≢ xor(exclusive or) addition
The following truth table defines a set of states of variables a,b,c,d.
Give a boolean expression in disjunctive normal form that is true in exactly
the states defined by the truth table. Based on this example, outline a
procedure that translates any such truth table into an equivalent boolean
expression in disjunctive normal form.
a  b  c  d
t  t  t  f
t  f  t  f
f  t  t  f
(a∧b∧c∧¬d)∨(a∧¬b∧c∧¬d)∨(¬a∧b∧c∧¬d)
In accordance with the topic at hand, we meticulously transform each 'false'
state into 'true' within the states that are stipulated to yield 'true' outputs.
Utilizing the Disjunctive Normal Form, an output of 'true' is generated provided
any condition is satisfied.
Thus, we construct a satisfying condition for each individual state.
3.89
The following truth table defines a set of states of variables a,b,c,d.
Give a boolean expression in conjunctive normal form that is true in exactly
the states defined by the truth table. Based on this example, outline a
procedure that translates any such truth table into an equivalent boolean
expression in conjunctive normal form.
Since every boolean expression can be described by such a truth table, every
boolean expression can be transformed to conjunctive normal form.
a  b  c  d
t  t  t  f 1110
t  t  t  t 1111
t  f  t  f 1010
f  t  t  f 0110
 a b c d
|1|1|0|1|¬a∨¬b∨c∨¬d
|1|1|0|0|¬a∨¬b∨c∨d
|1|0|1|1|¬a∨b∨¬c∨¬d
|1|0|0|1|¬a∨b∨c∨¬d
|1|0|0|0|¬a∨b∨c∨d
|0|1|1|1|a∨¬b∨¬c∨¬d
|0|1|0|1|a∨¬b∨c∨¬d
|0|1|0|0|a∨¬b∨c∨d
|0|0|1|1|a∨b∨¬c∨¬d
|0|0|1|0|a∨b∨¬c∨d
|0|0|0|1|a∨b∨c∨¬d
|0|0|0|0|a∨b∨c∨d

|a|b|c|d|
|1|1|1|1| X
|1|1|1|0| X
|1|1|0|1|
|1|1|0|0|
|1|0|1|1|
|1|0|1|0| X
|1|0|0|1|
|1|0|0|0|
|0|1|1|1|
|0|1|1|0| X
|0|1|0|1|
|0|1|0|0|
|0|0|1|1|
|0|0|1|0|
|0|0|0|1|
|0|0|0|0|
(¬a∨¬b∨c∨¬d)∧(¬a∨¬b∨c∨d)∧(¬a∨b∨¬c∨¬d)∧(¬a∨b∨c∨¬d)∧(¬a∨b∨c∨d)∧(a∨¬b∨¬c∨¬d)∧(a∨¬b∨c∨¬d)∧(a∨¬b∨c∨d)∧(a∨b∨¬c∨¬d)∧(a∨b∨¬c∨d)∧(a∨b∨c∨¬d)∧(a∨b∨c∨d)
In accordance with the topic at hand, we meticulously enumerate those states
that should yield a false output, derived inversely from the states that are
stipulated to produce a true output. We then proceed to methodically transform
each instance of truth within these states into falsehoods.
By harnessing the inherent properties of the conjunctive normal form,
which dictates that the output will be false should any single condition
remain unfulfilled, we generate a unique unsatisfied condition for each individual state.
4.1 p → (q → p)
4.2 Monotonicity of ∨: (p → q) → (p ∨ r → q ∨ r)
	p ∨ r → q ∨ r
	 =   <Def. of →(3.57), with p,q:=p∨r,q∨r —replace p∨r→q∨r
		  by p∨r∨q∨r≡q∨r>
		p ∨ r ∨ q ∨ r ≡ q ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=q,r —replace q∨r by r∨r>
		p ∨ r ∨ r ∨ q ≡ q ∨ r
	 =   <Idempotency of ∨(3.26), with p:=r —replace r∨r by r>
		p ∨ r ∨ q ≡ q ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=q,r —replace q∨r by r∨q>
		p ∨ r ∨ q ≡ r ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=p,r —replace p∨r by r∨p>
		r ∨ p ∨ q ≡ r ∨ q
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p∨q,q —replace
		  r∨p∨q≡r∨q by r∨(p∨q≡q)>
		r ∨ (p ∨ q ≡ q)
	 =   <Symmetry of ∨(3.24), with p,q:=r∨(p∨q≡q) —replace
		  r∨(p∨q≡q) by (p∨q≡q)∨r>
		(p ∨ q ≡ q) ∨ r
	 ⇐   <Weakening(3.76a), with p,q:=(p∨q≡q),r —replace (p∨q≡q)∨r by p∨q≡q>
		p ∨ q ≡ q
	 =   <Def. of →(3.57) —replace p∨q≡q by p→q>
		p → q
	p → q
	 =   <Def. of →(3.57) —replace p→q by p∨q≡q>
		p ∨ q ≡ q
	 →   <Weakening(3.76a), with q:=r —replace (p∨q≡q)∨r>
		(p ∨ q ≡ q) ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=(p∨q≡q),r —replace
		  (p∨q≡q)∨r by r∨(p∨q≡q)>
		r ∨ (p ∨ q ≡ q)
	 =   <Distributivity of ∨ over ≡(3.27), with p,q,r:=r,p∨q,q —replace
		  r∨(p∨q≡q) by r∨p∨q≡r∨q>
		r ∨ p ∨ q ≡ r ∨ q
	 =   <Symmetry of ∨(3.24), with p:=r —replace r∨q by q∨r>
		r ∨ p ∨ q ≡ q ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=r,p∨q —replace r∨p∨q by p∨q∨r>
		p ∨ q ∨ r ≡ q ∨ r
	 =   <Idempotency of ∨(3.26), with p:=r —replace r by r∨r>
		p ∨ q ∨ r ∨ r ≡ q ∨ r
	 =   <Symmetry of ∨(3.24), with p,q:=q,r —replace q∨r by r∨q>
		p ∨ r ∨ q ∨ r ≡ q ∨ r
	 =   <Def. of →(3.57) —replace p∨r∨q∨r≡q∨r by p∨r→q∨r>
		p ∨ r → q ∨ r
4.3 Monotonicity of ∧: (p → q) → (p ∧ r → q ∧ r)
4.4 (Extended) Deduction Theorem
	Suppose adding P1, …, Pn as axioms to propositional logic E, with the variables
	-of the Pi considered to be constants, allows Q to be proved.
	All the proofs in practical math rely solely on the deduction theorem.
	Proving p->q involves treating p as an additional axiom and then proving q.
	Then P1 ∧ … ∧ Pn → Q is a theorem.
	Proof.
		To prove p ∧ q → (p ≡ q), we assume the conjuncts of its antecedent and
		prove its consequent:
		Assume p, q
			p
		 =   <Assumption p>
			true
		 =   <Assumption q>
			q
	Proof.
		(p → p') → ((q → q') → (p ∧ q → p' ∧ q'))
		Assume p → p' (which is equivalent to p ∧ p' ≡ p)
		  Assume q → q' (which is equivalent to q ∧ q' ≡ q)
			p ∧ q
		  =   <Assumption p ∧ p' ≡ p>
			p ∧ p' ∧ q
		  =   <Assumption q ∧ q' ≡ q>
			p ∧ p' ∧ q ∧ q'
		  →   <Weakening(3.76b)>
			p' ∧ q'
Proof by case analysis:
	((a ∨ b ∨ c) ∧ (a → d) ∧ (b → d) ∧ (c → d)) → d
4.5 Metatheorem Case Analysis. If E[z:=true] and E[z:=false] are theorems, then so
	is E[z:=p].
	 E[z:=p]
	 =   <Shannon(3.89)>
		(p ∧ E[z:=true]) ∨ (¬p ∧ E[z:=false])
	 =   <Hypotheses of (4.5) together with Exercise 3.4>
		(p ∧ true) ∨ (¬p ∧ true)
	 =   <Identity of ∧(3.39), twice>
		p ∨ ¬p —Excluded Middle(3.28)
	We illustrate with two proofs of contradiction(3.42), p ∧ ¬p ≡ false;
	Proof. 
		If p is true, then the LHS of the formula is true ∧ ¬true, which,
		by Identity of ∧(3.39) and the Definition of false(3.8) is equivalent to false.
		If p is false, then the LHS of the formula is false ∧ ¬false, which by
		Zero of ∧(3.40) is equivalent to false. Hence, in both cases, the LHS is equivalent
		to false and the formula is true.
		Therefore, by metatheorem(4.5), the formula is true.
	Proof.
		Prove: p ∧ ¬p ≡ false
		 By Shannon
		 Case (p ∧ ¬p ≡ false)[p:=true]
		  =   <Textual substitution>
			true ∧ ¬true ≡ false
		  =   <Identity of ∧(3.39); Def. of false(3.8)>
			false ≡ false
		 Case (p ∧ ¬p ≡ false)[p:=false]
		  =   <Textual substitution>
			false ∧ ¬false ≡ false
		  =   <Zero of ∧(3.40)>
			false ≡ false —Reflexivity of ≡(3.5)
4.6 (p ∨ q ∨ r) ∧ (p → s) ∧ (q → s) ∧ (r → s) → s
	Fibonacci numbers f.i for i a nature number by
	f.i = { 0: if i = 0
			1: if i = 1
			f(i-1) + f(i-2): if i > 1 }
	Prove: S
	 By caseS: P,Q,R
		(proof of P ∨ Q ∨ R —omitted if obvious)
	 Case P: (proof of P → S)
	 Case Q: (proof of Q → S)
	 Case R: (proof of R → S)
Proof by mutual implication:
4.7 Proof method. To prove P ≡ Q, prove P → Q and Q → P.
4.8 Theorem. For any natural number i, even.i ≡ even(i²)
	even.i
	 =   <Def. of even>
		i = 2•k (for some natural number k)
	 =   <x = y ≡ x² = y²> (for nature numbers x,y)
		i² = (2•k)² (for some natural number k)
	 =   <Arithmetic>
		i² = 2•(2•k²) (for some natural number k)
	 =   <Definition of even>
		even(i²)
Proof by contradiction:
4.9 Proof by contradiction(3.74,3.12): ¬p → false ≡ p
4.10 Theorem. Let u be a left identity and v be a right identity of operator ∘,
	 i.e. u∘x=x and x∘v=x for all x. Then u=v.
	 Proof. We assume u ≠ v and prove a contradiction. Consider the expression
	 u ∘ v. Since u is a left identity, this expression equals v;
	 since v is a right identity, this expression equals u; hence, u=v,
	 but this contradicts the assumption u ≠ v. Hence the assumption is false,
	 and u = v.
	 u
	 =   <v is a right identity>
	 u ∘ v
	 =   <u is a left identity>
	 v
4.11 Theorem. Function Halt does not exist.
	 {Halt(P) ≡ P halts}
	 function Halt(P:string):bool;
	 begin … end
	 Proof. Assume Halt exists and consider the following procedure.
		procedure B;
		begin while Halt("call B") do skip end
	 We observe the following.
		"call B" halts
	 =   <Inspection of B's procedure body>
		¬Halt("call B")
	 =   <Definition of Halt —see comment on function Halt>
		¬("call B" halts)
	 We have derived a contradiction, so we have disproved the assumption that
	 Halt exists.
	 Assuming the existence of "duck," the duck function returns false if it
	 encounters code that causes it to hang, and true if it halts without issues.
	 The check function is designed to test "duck."
	 It establishes a while loop that continually prompts "duck" to inspect
	 the check function (leading to a hang).
	 We can observe that based on common knowledge, a while loop halts only when
	 the condition is false – that is, when "duck" returns false (encountering
	 code that causes it to hang). When "duck" reads the code, it recognizes
	 that the sole purpose of the check function is to call "duck."
	 Therefore, "duck" inspects its own code, and by definition, it doesn't hang.
	 As a result, "duck" concludes that "check" doesn't hang either,
	 given that "check" solely invokes "duck." This perpetual true evaluation
	 leads to an infinite loop within "check," causing it to hang indefinitely.
Proof by contrapositive
4.12 Proof method: Prove P → Q by proving its contrapositive ¬Q → ¬P (see (3.61)).
	 x + y ≥ 2 → x ≥ 1 ∨ y ≥ 1
	 By Contrapositive(3.61), De Morgan, and arithmetic, this formula is equivalent
	 to x < 1 ∧ y < 1 → x + y < 2
		x + y
	 <   <Assumptions x<1 and y<1>
		1+1
	 =   <Arithmetic>
		2
4.1 Prove theorem(4.1), p → (q → p), using the method of Sec 4.1.
	Answer:
	 q → p
	 =   <Def. of →(3.59), with p,q:=q,p —replace q→p by ¬q∨p>
		¬q ∨ p
	 =   <Symmetry of ∨(3.24), with p,q:=¬q,p —replace ¬q∨p by p∨¬q>
		p ∨ ¬q
	 ⇐   <Weakening(3.76a), with p,q:=p,¬q —replace p∨¬q by p>
		p
4.2 Prove Monotonicity of ∧(4.3), (p → q) → (p ∧ r → q ∧ r),
	using the method of sec 4.1. Start with the consequent,
	since it has more structure.
	Answer:
	 (p → q) → (p ∧ r → q ∧ r)
	 =   <Currying(3.65), with p,q,r:=(p→q),p∧r,q∧r —replace
		  (p→q)→(p∧r→q∧r) by (p→q)∧(p∧r)→q∧r>
		(p → q) ∧ (p ∧ r) → q ∧ r
	 (p → q) ∧ (p ∧ r)
	 =   <Associativity of ∧(3.37), with p,q:=(p→q),p —replace 
		(p→q)∧(p∧r) by ((p→q)∧p)∧r>
		((p → q) ∧ p) ∧ r
	 =   <Symmetry of ∧(3.36), with p,q:=(p→q),p —replace (p→q)∧p
		  by p∧(p→q)>
		(p ∧ (p → q)) ∧ r
	 →   <Modus ponens(3.77) —replace p∧(p→q) by q>
		q ∧ r
4.3 Prove Weakening/strengthening(3.76d), p ∨ (q ∧ r) → p ∨ q, using the method of Sec 4.1. Start
	with the antecedent, since it has more structure and distribute.
	 Answer:
	 true
	 =   <Leibniz(3.83), with e,f:=r,true>
		(r ≡ true) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ (p ∨ (q ∧ true)) → p ∨ q)
	 =   <Identity of ∧(3.39), with p:=q —replace q∧true by q>
		(r ≡ true) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ p ∨ q → p ∨ q)
	 =   <Metatheorem(3.7), Reflexivity of →(3.71), with p:=p∨q and True(3.4)
		  —replace p∨q→p∨q by true>
		(r ≡ true) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ true)
	 =   <Symmetry of ≡(3.2), with p,q:=(p∨(q∧r))→(p∨q),true —replace
		  (p∨(q∧r))→(p∨q)≡true by true≡(p∨(q∧r))→(p∨q)>
		(r ≡ true) → (true ≡ (p ∨ (q ∧ r)) → (p ∨ q))
	 =   <Identity of ≡(3.3), with q:=(p∨(q∧r))→(p∨q) —replace
		  true≡(p∨(q∧r))→(p∨q) by (p∨(q∧r))→(p∨q)>
		(r ≡ true) → (p ∨ (q ∧ r)) → (p ∨ q) —Lemma01
	 true
	 =   <Leibniz(3.83), with e,f:=r,false>
		(r ≡ false) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ (p ∨ (q ∧ false)) → p ∨ q)
	 =   <Zero of ∧(3.39), with p:=q —replace q∧false by false>
		(r ≡ false) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ p ∨ false → p ∨ q)
	 =   <Identity of ∨(3.30) —replace p∨false by p>
		(r ≡ false) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ p → p ∨ q)
	 =   <Metatheorem(3.7), Weakening→(3.76a) and True(3.4) —replace p→p∨q by true>
		(r ≡ false) → ((p ∨ (q ∧ r)) → (p ∨ q) ≡ true)
	 =   <Symmetry of ≡(3.2), with p,q:=(p∨(q∧r))→(p∨q),true —replace
		  (p∨(q∧r))→(p∨q)≡true by true≡(p∨(q∧r))→(p∨q)>
		(r ≡ false) → (true ≡ (p ∨ (q ∧ r)) → (p ∨ q))
	 =   <Identity of ≡(3.3), with q:=(p∨(q∧r))→(p∨q) —replace
		  true≡(p∨(q∧r))→(p∨q) by (p∨(q∧r))→(p∨q)>
		(r ≡ false) → (p ∨ (q ∧ r)) → (p ∨ q) —Lemma02
	 If Lemma01 and Lemma02 are theorems, then so is (3.76d).
4.4 Prove (p → q) ∧ (r → s) → (p ∨ r → q ∨ s), using the proof format of Sec. 4.1.
	You may first want to remove the implications in the antecedent, distribute
	as much as possible, and then use theorem(3.76d) and an absorption theorem.
	Answer:
	 (p → q) ∧ (r → s)
	 =   <Identity of ∧(3.39), with p:=(p→q) —replace (p→q) by (p→q)∧true>
		(p → q) ∧ true ∧ (r → s)
	 =   <Identity of ∧(3.39), with p:=(r→s) —replace (r→s) by (r→s)∧true>
		(p → q) ∧ true ∧ (r → s) ∧ true
	 =   <Metatheorem(3.7), True(3.4) and Weakening(3.76a) —replace true
		  by (q→q∨s)>
		(p → q) ∧ (q → q ∨ s) ∧ (r → s) ∧ true
	 =   <Metatheorem(3.7), True(3.4) and Weakening(3.76a) —replace true
		  by (s→q∨s)>
		(p → q) ∧ (q → q ∨ s) ∧ (r → s) ∧ (s → q ∨ s)
	 →   <Transitivity(3.82a), with r:=q∨s and Monotonicity of ∧(4.3),
		  with p,q,r:=(p→q),(q→q∨s),(r→s)∧(s→q∨s) —replace (p→q)∧(q→q∨s)
		  by (p→q∨s)>
		(p → q ∨ s) ∧ (r → s) ∧ (s → q ∨ s)
	 =   <Symmetry of ∧(3.36), with p,q:=(p→q∨s),(r→s)∧(s→q∨s) —replace
		  (p→q∨s)∧(r→s)∧(s→q∨s) by (r→s)∧(s→q∨s)∧(p→q∨s)>
		(r → s) ∧ (s → q ∨ s) ∧ (p → q ∨ s)
	 →   <Transitivity(3.82a), with p,q,r:=r,s,q∨s and Monotonicity of ∧(4.3),
		  with p,q,r:=(r→s),(s→q∨s),(p→q∨s) —replace (r→s)∧(s→q∨s)
		  by (r→q∨s)>
		(r → q ∨ s) ∧ (p → q ∨ s)
	 =   <Symmetry of ∧(3.36), with p,q:=(r→q∨s),(p→q∨s) —replace (r→q∨s)∧(p→q∨s)
		  by (p→q∨s)∧(r→q∨s)>
		(p → q ∨ s) ∧ (r → q ∨ s)
	 =   <(3.78), with q,r:=r,q∨s —replace (p→q∨s)∧(r→q∨s) by p∨r→q∨s>
		p ∨ r → q ∨ s
4.5 Prove (p → q) ∧ (r → s) → (p ∧ r → q ∧ s), using the proof format of Sec. 4.1.
	Before using the proof format, you may first want to using Currying(3.65)
	to move p ∧ r into the antecedent.
	Answer:
	 (p → q) ∧ (r → s) → (p ∧ r → q ∧ s)
	 =   <Currying(3.65), with p,q,r:=(p→q)∧(r→s),p∧r,q∧s —replace
		  (p→q)∧(r→s)→(p∧r→q∧s) by (p→q)∧(r→s)∧p∧r→q∧s>
		(p → q) ∧ (r → s) ∧ p ∧ r → q ∧ s
	 (p → q) ∧ (r → s) ∧ p ∧ r
	 =   <Symmetry of ∧(3.36), with p,q:=(p→q)∧(r→s),p —replace (p→q)∧(r→s)∧p
		  by p∧(p→q)∧(r→s)>
		p ∧ (p → q) ∧ (r → s) ∧ r
	 =   <Symmetry of ∧(3.36), with p,q:=(r→s),r —replace (r→s)∧r by r∧(r→s)>
		p ∧ (p → q) ∧ r ∧ (r → s)
	 →   <Modus ponens(3.77) —replace p∧(p→q) by q>
		q ∧ r ∧ (r → s)
	 →   <Modus ponens(3.77), with p,q:=r,s —replace r∧(r→s) by s>
		q ∧ s
4.6 Prove p → (q → p) by the method of assuming the antecedent.
	Answer:
	Assume p
	 q → p
	 =   <Assumption>
		q → true
	 =   <Right zero of →(3.72), with p:=q —replace q→true by true>
		true
4.7 Prove (¬p → q) → ((p → q) → q) by the method of assuming the antecedent.
	Answer:
	Assume ¬p → q
	  Assume p → q
		q
	  =   <(3.79), with r:=q —replace q by (p→q)∧(¬p→q)>
		(p → q) ∧ (¬p → q)
	  =   <Assumption>
		true ∧ true
4.8 Prove p ∧ q → (p ≡ q) by the method of assuming the antecedent.
	Answer:
	Assume p, q
		p
	 =   <Assumption p>
		true
	 =   <Assumption q>
		q
4.9 Prove (p → p') ∧ (q → q') → (p ∨ q → p' ∨ q') by the method of assuming the
	antecedent.
	Answer:
	Assume (p → p')(which is equivalent to p ∧ p' ≡ p), 
		   (q → q')(which is equivalent to q ∧ q' ≡ q)
		p ∨ q
	 =   <Assumption p ∧ p' ≡ p>
		p ∧ p' ∨ q
	 =   <Assumption q ∧ q' ≡ q>
		p ∧ p' ∨ q ∧ q'
	 =   <Symmetry of ∧(3.36), with p,q:=p,p' —replace p∧p' by p'∧p>
		p' ∧ p ∨ q ∧ q'
	 =   <Symmetry of ∧(3.36), with p,q:=q,q' —replace q∧q' by q'∧q>
		p' ∧ p ∨ q' ∧ q
	 →   <Weakening(3.76b), with p,q:=p',p —replace p'∧p by p'>
		p' ∨ q' ∧ q
	 →   <Weakening(3.76b), with p,q:=q',q —replace q'∧q by q'>
		p' ∨ q'
4.10 Prove Modus ponens(3.77), p ∧ (p → q) → q, by the method of assuming the antecedent.
	Answer:
	Assume p,(p → q) (which is equivalent to p ∨ q ≡ q)
		q
	 =   <Assumption p ∨ q ≡ q>
		p ∨ q
	 =   <Assumption p>
		true ∨ q
	 =   <Symmetry of ∨(3.24), with p,q:=true,q —replace true∨q
		  by q∨true>
		q ∨ true
	Proof.
		Prove: p ∧ ¬p ≡ false
		 By Shannon
		 Case (p ∧ ¬p ≡ false)[p:=true]
		  =   <Textual substitution>
			true ∧ ¬true ≡ false
		  =   <Identity of ∧(3.39); Def. of false(3.8)>
			false ≡ false
		 Case (p ∧ ¬p ≡ false)[p:=false]
		  =   <Textual substitution>
			false ∧ ¬false ≡ false
		  =   <Zero of ∧(3.40)>
			false ≡ false —Reflexivity of ≡(3.5)
4.11 Prove the following theorem using Metatheorem Case analysis(4.5):
	(p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r)
	Answer:
	Prove: (p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r)
	 By Shannon
	 Case r=true:
	 ((p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r))[r:=true]
	 =   <Textual substitution>
		(p ∨ q) ∧ true ≡ (p ∧ true) ∨ (q ∧ true)
	 =   <Identity of ∧(3.39), with p:=(p∨q) —replace (p∨q)∧true
		  by p∨q>
		p ∨ q ≡ (p ∧ true) ∨ (q ∧ true)
	 =   <Identity of ∧(3.39) —replace p∧true by p>
		p ∨ q ≡ p ∨ (q ∧ true)
	 =   <Identity of ∧(3.39), with p:=q —replace q∧true by q>
		p ∨ q ≡ p ∨ q —Reflexivity of ≡(3.5)
	 Case r=false:
	 ((p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r))[r:=false]
	 =   <Textual substitution>
		(p ∨ q) ∧ false ≡ (p ∧ false) ∨ (q ∧ false)
	 =   <Zero of ∧(3.40), with p:=(p∨q) —replace (p∨q)∧false by false>
		false ≡ (p ∧ false) ∨ (q ∧ false)
	 =   <Zero of ∧(3.40) —replace p∧false by false>
		false ≡ false ∨ (q ∧ false)
	 =   <Zero of ∧(3.40), with p:=q —replace q∧false by false>
		false ≡ false ∨ false
	 =   <Identity of ∨(3.30), with p:=false —replace false∨false by false>
		false ≡ false —Reflexivity of ≡(3.5)
4.12 Prove theorem(4.6), (p ∨ q ∨ r) ∧ (p → s) ∧ (q → s) ∧ (r → s) → s
	 Answer:
	 Prove: (p ∨ q ∨ r) ∧ (p → s) ∧ (q → s) ∧ (r → s) → s
	 By Shannon
	 Case s=true:
	 ((p ∨ q ∨ r) ∧ (p → s) ∧ (q → s) ∧ (r → s) → s)[s:=true]
	 =   <Textual substitution>
		(p ∨ q ∨ r) ∧ (p → true) ∧ (q → true) ∧ (r → true) → true
	 =   <Right zero of →(3.72), with p:=(p∨q∨r)∧(p→true)∧(q→true)∧(r→true) —replace
		  (p∨q∨r)∧(p→true)∧(q→true)∧(r→true)→true by true>
		true
	 Case s=false:
	 ((p ∨ q ∨ r) ∧ (p → s) ∧ (q → s) ∧ (r → s) → s)[s:=false]
	 =   <Textual substitution>
		(p ∨ q ∨ r) ∧ (p → false) ∧ (q → false) ∧ (r → false) → false
	 =   <(3.74) four times>
		¬((p ∨ q ∨ r) ∧ ¬p ∧ ¬q ∧ ¬r)
	 =   <De Morgan(3.47b) twice>
		¬((p ∨ q ∨ r) ∧ ¬(p ∨ q ∨ r))
	 =   <Contradiction(3.42)>
		¬false
	 =   <Negation of false(3.13)>
		true
4.13 let x↓y be the minimum of integers x and y, defined by x↓y=(if x ≤ y then x
	 else y). Prove that ↓ is symmetric, i.e. b↓c=c↓b. How many cases do you have
	 to consider? You may use the necessary rules of integer arithmetic, for example,
	 that b ≤ c ≡ b = c ∨ b < c and that b < c ≡ c > b.
	 Answer:
	 Three cases.
		Note: 
		b ≤ c ≡ (b = c) ∨ (b < c) and that b < c ≡ c > b
	 x↓y = {x: if x = y,
			x: if x < y,
			y: if x > y,}
	 if x=y then min(x,y)=(x<=y?x:y)=x=y=(y<=x?y:x)=min(y,x)
	 if x<y then min(x,y)=(x<=y?x:y)=x=(y<=x?y:x)=min(y,x)
	 if x>y then min(x,y)=(x<=y?x:y)=y=(y<=x?y:x)=min(y,x)
4.14 Prove by case analysis that ↓ is associative, i.e. that b↓(c↓d)=((b↓c)↓d)
	 (see the previous exercise). How many cases do you have to consider, based
	 on the definition of ↓?
	 Answer:
	 3! (a,b,c permutation)
		a__
			ab_
				abc
			ac_
				acb
		b__
			ba_
				bac
			bc_
				bca
		c__
			ca_
				cab
			cb_
				cba
	 x↓(y↓z) = {(a: if a≤b≤c){
					a: a<b<c,
					a: a<b=c,
					a: a=b<c,
					a: a=b=c,
					},
				(a: if a≤c≤b){
					a: a<c<b,
					a: a<c=b,
					a: a=c<b,
					a: a=c=b,
					},
				(b: if b≤a≤c){
					b: b<a<c,
					b: b<a=c,
					b: b=a<c,
					b: b=a=c,
					},
				(b: if b≤c≤a){
					b: b<c<a,
					b: b<c=a,
					b: b=c<a,
					b: b=c=a,
					},
				(c: if c≤a≤b){
					c: c<a<b,
					c: c<a=b,
					c: c=a<b,
					c: c=a=b,
					},
				(c: if c≤b≤a){
					c: c<b<a,
					c: c<b=a,
					c: c=b<a,
					c: c=b=a,
					},
				}
		Note: 
		b ≤ c ≡ (b = c) ∨ (b < c) and that b < c ≡ c > b
			Following the logic from the previous question, each possibility
			can be divided into four categories:
			b<c<d, b<=c<d, b<c<=d, and b=c=d.
			In total, there are 6*4=24 different scenarios. However,
			cases like b=c=d and c=b=d are equivalent, resulting in 19 cases.
	 Having approached symmetry and associativity as distinct concepts,
	 we've employed different methods for each, resulting in six distinct cases
	 that have already been demonstrated.
	 if a<=b<=c then min(a,min(b,c))=min(a,b)=a=min(a,c)=min(min(a,b),c)
	 if a<=c<=b then min(a,min(c,b))=min(a,c)=a=min(a,b)=min(min(a,c),b)
	 if b<=a<=c then min(b,min(a,c))=min(b,a)=b=min(b,c)=min(min(b,a),c)
	 if b<=c<=a then min(b,min(c,a))=min(b,c)=b=min(b,a)=min(min(b,c),a)
	 if c<=a<=b then min(c,min(a,b))=min(c,a)=c=min(c,b)=min(min(c,a),b)
	 if c<=b<=a then min(c,min(b,a))=min(c,b)=c=min(c,a)=min(min(c,b),a)
4.15 Consider the discussion on page 76 that shows how a proof of P≡Q with P∨Q as
	 an intermediate step can be viewed as a proof by mutual implication.
	 Write a similar discussion to show how a proof of P≡Q with P∧Q as an intermediate
	 step can be viewed as a proof by mutual implication.
	 Answer:
	 P
	 =   <Hint>
		…
	 =   <Hint>
		P ∧ Q
	 =   <Hint>
		…
	 =   <Hint>
	 Q
	 This proof establishes (P≡P∧Q) and (P∧Q≡Q). Since P≡P∧Q equivales P→Q
	 and P∧Q≡Q equivales Q→P, the proof establishes (P→Q)∧(Q→P).
	 But this formula is the LHS of (3.80). Hence, the proof is really just
	 a proof by mutual implication of P→Q.
5.1 If Joe failes to submit a project in course CS414, then he fails the course.
	If Joe fails CS414, then he cannot graduate. Hence, if Joe graduates,
	he must have submitted a project.
		F0: If Joe failes to submit a project in course CS414,
			then he fails the course.
		F1: If Joe fails CS414, then he cannot graduate.
		C: Hence, if Joe graduates, he must have submitted a project.
	F0 ∧ F1 → C
		s: Joe submits a project in CS414.
		f: Joe fails CS414.
		g: Joe graduates.
	F0= ¬s→f
	F1= f→¬g
	C= g→s
		(¬s → f) ∧ (f → ¬g) → (g → s)
	Proof.
		(¬s → f) ∧ (f → ¬g)
	 →   <Transitivity(3.82a)>
		¬s → ¬g
	 =   <Contrapositive(3.61)>
		g → s
5.2 If X is greater than zero, then if Y is zero than Z is zero.
	Variable Y is zero. Hence, either X is greater than zero or Z is zero.
		x: X is greater than zero.
		y: Y is zero.
		z: Z is zero.
5.3 (x → (y → z)) ∧ y → x ∨ z
		(x → (y → z)) ∧ y
	 =   <Currying(3.65) twice>
		(y → (x → z)) ∧ y
	 =   <(3.66)>
		(x → z) ∧ y
	strcmp("(x→z)∧y","x∨z")≠0
	Variable y has nothing to do with the consequent, and x → z (i.e. (¬x ∨ z))
	does not imply x ∨ z. Hence, we should suspect that (5.3) is not valid and
	that argument (5.2) is not sound.
	Counterexamples for Expressions
	expression|counterexample 1|counterexample 2|
		   p∧q|         p=false|         q=false|
		   p∨q|       p=q=false|                |
		   p≡q| p=true, q=false| p=false, q=true|
		   p≢q|        p=q=true|       p=q=false|
		   p→q|p=true, q =false|                |
	Counterexample of 5.3:
	x = z = false and y = true
		X is not greater than zero
		Z is not zero
		and
		Y is zero.
Solving Puzzles:
	Portia's suitor's dilemma
	 Portia has a gold casket and a silver casket and has placed a picture of herself
	 in one of them. On the caskets, she has written the following inscriptions:
		Gold: The portrait is not in here.
		Silver: Exactly one of these inscriptions is true.
	 Portia explains to her suitor that each inscription may be true or false,
	 but that she has placed her portrait in one of the caskets in a manner that is
	 consistent with this truth or falsity of the inscriptions. If she can choose
	 the casket with her portrait, she will marry her —in those days, that's what
	 suitors wanted. The problem for the suitor is to use the inscriptions (although
	 they could be true or false) to determine which casket contains her portrait.
	 gc: The portrait is in the gold casket.
	 sc: The portrait is in the silver casket.
	 g: The portrait is not in the gold casket.
	 s: Exactly one of g and s is true.
	 F0: gc ≡ ¬sc
	 F1: g ≡ ¬gc
	 F2: s ≡ (s ≡ ¬g)
	 Proof.
		s ≡ (s ≡ ¬g)
	 =   <Symmetry of ≡(3.2)>
		¬g
	 =   <F1; Double negation(3.12)>
		gc
	Hence, from F1 and F2 (F0 is not needed) we conclude gc. The portrait is in
	the gold casket.
Solving Puzzles:
	Superman
	 If Superman were able and willing to prevent evil, she would do so. If Superman
	 were unable to prevent evil, she would be impotent; if she were unwilling to
	 prevent evil, she would be malevolent. Superman does not prevent evil. If
	 Superman exists, she is neither impotent nor malevolent. Therefore, Superman
	 does not exist.
		a: Superman is able to prevent evil.
		w: Superman is willing to prevent evil.
		i: Superman is impotent.
		m: Superman is malevolent.
		p: Superman prevents evil.
		e: Superman exists.
		F0: a ∧ w → p
		F1: (¬a → i) ∧ (¬w → m)
		F2: ¬p
		F3: e → ¬i ∧ m
5.4 F0 ∧ F1 ∧ F2 ∧ F3 → ¬e
	Proof.
	Assume F0, F1, F2, F3
		¬e
	 ⇐   <Contrapositive ¬(¬i ∧ ¬m) → ¬e of F3
		  the only other place e appears>
		¬(¬i ∧ ¬m)
	 =   <De Morgan(3.47a); Double negation(3.12), twice>
		i ∨ m
	 ⇐   <First conjunct of F1 and Monotonicity(4.2)>
		¬a ∨ m
	 ⇐   <Second conjunct of F1 and Monotonicity(4.2)>
		¬w ∨ ¬a
	 =   <De Morgan(3.47a)>
		¬(a ∧ w)
	 ⇐   <Contrapositive ¬p → ¬(a ∧ w) of F0>
		¬p —this is F2
		Note:
		(¬a → i) → (¬a ∨ m → i ∨ m)
		(¬w → m) → (¬w ∨ ¬a → m ∨ ¬a)
5.5 c = (a ∧ b) ∧
	𝜙 = ¬b ∧ 𝜃 = (a ∧ 𝜙) ∧
	𝜔 = ¬a ∧ 𝜋 = (b ∧ 𝜔) ∧
	s = (𝜃 ∨ 𝜔)
5.6 z1 = E1(x) ∧ z2 = E2(x) ∧ … ∧ zm = Em(x)
5.7 NOT(a): ¬a
	AND(a1,…,an): a1 ∧ … ∧ an (for n≥2)
	OR(a1,…an): a1 ∨ … ∨ an (for n≥2)
5.8 Definition. Circuit 
5.1 Formalize the following arguments and either prove that they are valid or find
	a counterexample.
	 (a)
		Either the program does not terminate or n eventually becomes 0.
		If n becomes 0, m will eventually be 0.
		The program terminates.
		Therefore, m will eventually be 0.
		Answer:
			t: The program does not terminates.
			n: n eventually becomes 0.
			m: m will eventually be 0.
			F0: t ≡ ¬n
			F1: n → m
			F2: ¬t
			C: m
			F0 ∧ F1 ∧ F2 → C
			Proof.
				Assume F0, F1, F2
					m
				 ⇐   <Assumption F1>
					n
				 =   <Assumption F0, Double negation(3.12)>
					¬t —This is F2
			Prove: (t ≡ ¬n) ∧ (n → m) ∧ ¬t → m
			By Case
			Case t=true:
				Case n=true:
					Case m=true:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=true,true,true]
					 =   <Textual Evaluation>
						((true ≡ ¬true) ∧ (true → true) ∧ ¬true → true)
					 =   <Boolean Expression Evaluation>
						true
					Case m=false:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=true,true,false]
					 =   <Textual Evaluation>
						((true ≡ ¬true) ∧ (true → false) ∧ ¬true → false)
					 =   <Boolean Expression Evaluation>
						true
				Case n=false:
					Case m=true:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=true,false,true]
					 =   <Textual Evaluation>
						((true ≡ ¬false) ∧ (false → true) ∧ ¬true → true)
					 =   <Boolean Expression Evaluation>
						true
					Case m=false:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=true,false,false]
					 =   <Textual Evaluation>
						((true ≡ ¬false) ∧ (false → false) ∧ ¬true → false)
					 =   <Boolean Expression Evaluation>
						true
			Case t=false:
				Case n=true:
					Case m=true:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=false,true,true]
					 =   <Textual Evaluation>
						((false ≡ ¬true) ∧ (true → true) ∧ ¬false → true)
					 =   <Boolean Expression Evaluation>
						true
					Case m=false:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=false,true,false]
					 =   <Textual Evaluation>
						((false ≡ ¬true) ∧ (true → false) ∧ ¬false → false)
					 =   <Boolean Expression Evaluation>
						true
				Case n=false:
					Case m=true:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=false,false,true]
					 =   <Textual Evaluation>
						((false ≡ ¬false) ∧ (false → true) ∧ ¬false → true)
					 =   <Boolean Expression Evaluation>
						true
					Case m=false:
						((t ≡ ¬n) ∧ (n → m) ∧ ¬t → m)[t,n,m:=false,false,false]
					 =   <Textual Evaluation>
						((false ≡ ¬false) ∧ (false → false) ∧ ¬false → false)
					 =   <Boolean Expression Evaluation>
						true
	 (b)
		If the initialization is correct and if the loop terminates, then P is true in the final state.
		P is true in the final state.
		Therefore, if the initialization is correct, the loop terminates.
		Answer:
			i: The initialization is correct.
			l: The loop terminates.
			p: P is true in the final state.
			F0: i ∧ l → p
			F1: p
			C: i → l
			F0 ∧ F1 → C
			Counterexample:
				(i,p,l:=true,true,false): F0,F1,C:=true,true,false
	 (c)
		If there is a man on the moon, the moon is made of cheese, and if the moon is made of cheese then I am a monkey.
		Either no man is on the moon or the moon is not made of cheese.
		Therefore either the moon is not made of cheese or I am a monkey.
		Answer:
			a: A man on the moon.
			m: The moon is made of cheese.
			i: I am a monkey.
			F0: (a → m) ∧ (m → i)
			F1: a ≡ ¬m
			C:  m ≡ i
			F0 ∧ F1 → C
			Proof.
				Assume F1
					(a → m) ∧ (m → i)
				 =   <Assumption>
					(¬m → m) ∧ (m → i)
				 =   <Def. of →(3.59)>
					(m ∨ m) ∧ (m → i)
				 =   <Idempotency of ∨(3.26)>
					m ∧ (m → i)
				 →   <Weakening(3.76a), Monotonicity of ∧(4.3)>
					(i → m) ∧ (m → i)
				 =   <Antisymmetry(3.81)>
					i ≡ m
				Note:
					m → (m → i)
					Weakening(3.76a), with p:=¬i
					(m ∨ ¬i) ∧ (m → i)
					= (i → m) ∧ (m → i)
				Prove: ((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i)
				By Case
				Case a=true:
					Case m=true:
						Case i=true:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=true,true,true]
						 =   <Textual Substitution>
							(((true → true) ∧ (true → true)) ∧ (true ≡ ¬true) → (true ≡ true))
						 =   <Boolean Expression Evaluation>
							true
						Case i=false:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=true,true,false]
						 =   <Textual Substitution>
							(((true → true) ∧ (true → false)) ∧ (true ≡ ¬true) → (true ≡ false))
						 =   <Boolean Expression Evaluation>
							true
					Case m=false:
						Case i=true:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=true,false,true]
						 =   <Textual Substitution>
							(((true → false) ∧ (false → true)) ∧ (true ≡ ¬false) → (false ≡ true))
						 =   <Boolean Expression Evaluation>
							true
						Case i=false:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=true,false,false]
						 =   <Textual Substitution>
							(((true → false) ∧ (false → false)) ∧ (true ≡ ¬false) → (false ≡ false))
						 =   <Boolean Expression Evaluation>
							true
				Case a=false:
					Case m=true:
						Case i=true:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=false,true,true]
						 =   <Textual Substitution>
							(((false → true) ∧ (true → true)) ∧ (false ≡ ¬true) → (true ≡ true))
						 =   <Boolean Expression Evaluation>
							true
						Case i=false:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=false,true,false]
						 =   <Textual Substitution>
							(((false → true) ∧ (true → false)) ∧ (false ≡ ¬true) → (true ≡ false))
						 =   <Boolean Expression Evaluation>
							true
					Case m=false:
						Case i=true:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=false,false,true]
						 =   <Textual Substitution>
							(((false → false) ∧ (false → true)) ∧ (false ≡ ¬false) → (false ≡ true))
						 =   <Boolean Expression Evaluation>
							true
						Case i=false:
							(((a → m) ∧ (m → i)) ∧ (a ≡ ¬m) → (m ≡ i))[a,m,i:=false,false,false]
						 =   <Textual Substitution>
							(((false → false) ∧ (false → false)) ∧ (false ≡ ¬false) → (false ≡ false))
						 =   <Boolean Expression Evaluation>
							true
	 (d)
		If Joe loves Mary, then either mom is mad or father is sad.
		Father is sad.
		Therefore, if mom is mad then Joe doesn't love Mary.
		Answer:
			j: Joe loves Mary.
			m: Mom is mad.
			d: Dad is sad.
			F0: j → (m ≡ ¬d)
			F1: d
			C: m → ¬j
			F0 ∧ F1 → C
			Proof.
				Assume F1
					(j → (m ≡ ¬d))
				 =   <Assumption F1>
					(j → (m ≡ ¬true))
				 =   <Def. of false(3.8)>
					(j → (m ≡ false))
				 =   <(3.15)>
					j → ¬m
				 =   <Contrapositive(3.61), Double negation(3.12)>
					m → ¬j
			Prove: (j → (m ≡ ¬d)) ∧ d → (m → ¬j)
			By Case
			Case j=true:
				Case m=true:
					Case d=true:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=true,true,true]
					 =   <Textual Substitution>
						(true → (true ≡ ¬true)) ∧ true → (true → ¬true)
					 =   <Boolean Expression Evaluation>
						true
					Case d=false:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=true,true,false]
					 =   <Textual Substitution>
						(true → (true ≡ ¬false)) ∧ false → (true → ¬true)
					 =   <Boolean Expression Evaluation>
						true
				Case m=false:
					Case d=true:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=true,false,true]
					 =   <Textual Substitution>
						(true → (false ≡ ¬true)) ∧ true → (false → ¬true)
					 =   <Boolean Expression Evaluation>
						true
					Case d=false:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=true,false,false]
					 =   <Textual Substitution>
						(true → (false ≡ ¬false)) ∧ false → (false → ¬true)
					 =   <Boolean Expression Evaluation>
						true
			Case j=false:
				Case m=true:
					Case d=true:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=false,true,true]
					 =   <Textual Substitution>
						(false → (true ≡ ¬true)) ∧ true → (true → ¬false)
					 =   <Boolean Expression Evaluation>
						true
					Case d=false:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=false,true,false]
					 =   <Textual Substitution>
						(false → (true ≡ ¬false)) ∧ false → (true → ¬false)
					 =   <Boolean Expression Evaluation>
						true
				Case m=false:
					Case d=true:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=false,false,true]
					 =   <Textual Substitution>
						(false → (false ≡ ¬true)) ∧ true → (false → ¬false)
					 =   <Boolean Expression Evaluation>
						true
					Case d=false:
						((j → (m ≡ ¬d)) ∧ d → (m → ¬j))[j,m,d:=false,false,false]
					 =   <Textual Substitution>
						(false → (false ≡ ¬false)) ∧ false → (false → ¬false)
					 =   <Boolean Expression Evaluation>
						true
5.2 Prove that the following argument is valid if the "or" in it is considered to
	be inclusive and invalid if it is considered exclusive.
	 If an algorithm is reliable, then it is okay. Therefore, either an algorithm
	 is okay or it is unreliable.
	Answer:
		r: An algorithm is reliable.
		o: An algorithm is okay.
		F0: r → o
		C0: r ≡ o
		C1: ¬r ≡ o
		Proof.
			Assume F0
				r ≡ o
			 ⇐   <Assumption>
				r ≡ r
			 =   <Reflexivity of ≡(3.5)>
				true
		Prove: ¬r ≡ o
		By case
		Case r=true:
			Case o=true:
				(¬r ≡ o)[r,o:=true,true]
			 =   <Textual Substitution>
				(¬true ≡ true)
			 =   <Boolean Expression Evaluation>
				false
			Case o=false:
				(¬r ≡ o)[r,o:=true,false]
			 =   <Textual Substitution>
				(¬true ≡ false)
			 =   <Boolean Expression Evaluation>
				true
		Case r=false:
			Case o=true:
				(¬r ≡ o)[r,o:=false,true]
			 =   <Textual Substitution>
				(¬false ≡ true)
			 =   <Boolean Expression Evaluation>
				true
			Case o=false:
				(¬r ≡ o)[r,o:=false,false]
			 =   <Textual Substitution>
				(¬false ≡ false)
			 =   <Boolean Expression Evaluation>
				false
5.3 Suppose we have the following facts. Prove that if the maid told the truth,
	the butler lied.
	 The maid said she saw the butler in the living room.
	 The living room adjoins kitchen.
	 The shot was fired in the kitchen and could be heard in all adjoining rooms.
	 The butler, who had good hearing, said he did not heard the shot.
	 Answer:
		m: The maid told the truth.
		b: The butler told the truth.
		l: The butler was in the living room.
		k: The living room adjoins the kitchen.
		h: The butler heard the shot.
		F0: m → l
		F1: k
		F2: (k ∧ l) → h
		F3: b → ¬h
		C: m → ¬b
		F0 ∧ F1 ∧ F2 ∧ F3 → C
		Proof.
			Assume F0, F1, F2, F3
				m
			 →   <Assumption F0>
				l
			 =   <Metatheorem(3.7), Identity of ∧(3.39) and Assumption F2, F2>
				l ∧ k ∧ (k ∧ l → h)
			 →   <Modus Ponens(3.77)>
				h
			 =   <Metatheorem(3.7), Identity of ∧(3.39), Assumption F3>
				h ∧ (b → ¬h)
			 =   <Contrapositive(3.61), Double negation(3.12)>
				h ∧ (h → ¬b)
			 →   <Modus Ponens(3.77)>
				¬b
6.1 Scope rule. For k an integer, line k of a proof may contain reference to lines
	1 through k-1. A line numbered …j.k (say) can reference lines …j.1 through
	j.(k-1) as well as whatever can be referenced by line …j.
6.2 p → (q → r) ⊢ p ∧ q → r
	1 p ∧ q ⊢ r
	  1.1 p      ∧–E, pr 1.a
	  1.2 q → r  →–E, 1.1, pr a
	  1.3 q      ∧–E, pr 1.a
	  1.4 r      →–E, 1.3, 1.2
	2 p ∧ q → r  →–I, 1
6.3 p ∧ q → r ⊢ p → (q → r)
	1 p ⊢ q → r
	  1.1 q ⊢ r
		  1.1.1 p ∧ q  ∧–I, pr 1.a, pr1.1.a
		  1.1.2 r      →–E, 1.1.1, pr a
	  1.2 q → r        →–I, 1.1
	2 p → (q → r)      →–I, 1
6.4 ⊢ p ≡ ¬¬p
	1 p ⊢ ¬¬p
	  1.1 ¬p ⊢ p ∧ ¬p
		  1.1.1 p ∧ ¬p   ∧–I, pr 1.a, pr 1.1.a
	  1.2 ¬¬p			 ¬–I, 1.1
	2 p → ¬¬p			 →–I, 1
	3 ¬¬p ⊢ p
	  3.1 ¬p ⊢ ¬p ∧ ¬¬p
		  3.1.1 ¬p ∧ ¬¬p ∧–I, pr 3.1.a, pr 3.a
	  3.2 p 		 	 ¬–E, 3.1
	4 ¬¬p → p			 →–I, 3
	5 p ≡ ¬¬p			 ≡–I, 2, 4
6.4 ⊢ p ≡ ¬¬p
	p 𠪪p by Mutual implication
	1 p → ¬¬p by Deduction
	  1.1 p Assumption
	  1.2 ¬¬p by Contradiction
		  1.2.1 ¬p Assumption
		  1.2.2 p ∧ ¬p ∧–I, 1.1, 1.2.1
	2 ¬¬p → p by Deduction
	  2.1 ¬¬p Assumption
	  2.2 p by Contradiction
		  2.2.1 ¬p Assumption
		  2.2.2 ¬p ∧ ¬¬p ∧–I, 2.2.1, 2.1
6.1 Using the example on page 109 of an equational proof and its Hilbert-style
	counterpart, give a procedure to transform any equational proof into a
	Hilbert-style proof.
	Equational:
		¬false
	 =   <Def. of false(3.8)>
		¬¬true
	 =   <Double negation(3.12)>
		true
	Hilbert-style:
	¬false ≡ true
	1 ¬false ≡ ¬¬true  Leibniz, (3.8)
	2 ¬¬true ≡ true    Substitution, (3.12)
	3 ¬false ≡ true    Transitivity (1.4), 1, 2
	假如要從以等式為基礎的證明轉換至希爾伯特的證明方式
	必須要進行以下步驟:
	skip for now
6.2 Prove p → (q → p)
	p → (q → p) by Deduction
	1 p Assumption
	2 q → p by Deduction
	  2.1 q Assumption 2
	  2.2 p 1
6.3 Prove p → (q → p ∧ q)
	p → (q → p ∧ q) by Deduction
	1 p Assumption
	2 q → p ∧ q by Deduction
	  2.1 q Assumption 2
	  2.2 p ∧ q ∧–I, 1, 2.1
6.4 Prove (p ∧ q) ∧ r → p ∧ (q ∧ r)
	(p ∧ q) ∧ r → p ∧ (q ∧ r) by Deduction
	1 p ∧ q ∧–E, Assumption
	2 r ∧–E, Assumption
	3 p ∧ (q ∧ r) by Deduction
	  3.1 p ∧–E 1
	  3.2 q ∧–E 1
	  3.3 q ∧ r ∧–I, 3.2, 2
6.5 Prove (p → q) → (p ∧ r → q ∧ r)
	(p → q) → (p ∧ r → q ∧ r) by Deduction
	1 p → q Assumption
	2 p ∧ r → q ∧ r by Deduction
	  2.1 p ∧–E, Assumption 2
	  2.2 r ∧–E, Assumption 2
	  2.3 q →–E, 2.1, 1
	  2.4 q ∧ r ∧–I, 2.3, 2.2
6.6 Prove (p → (q → r)) → (q → (p → r))
	(p → (q → r)) → (q → (p → r)) by Deduction
	1 p → (q → r) Assumption
	2 q → (p → r) by Deduction
	  2.1 q Assumption 2
	  2.2 p → r By deduction
		  2.2.1 p Assumption 2.2
		  2.2.2 r →–E, 2.2, 2.2.1
6.7 Prove (p → q) ∧ (p → r) → (p → q ∧ r)
	(p → q) ∧ (p → r) → (p → q ∧ r) by Deduction
	1 p → q ∧–E, Assumption
	2 p → r ∧–E, Assumption
	3 p → q ∧ r by Deduction
	  3.1 p Assumption 3
	  3.2 q →–E, 3.1, 1
	  3.3 r →–E, 3.1, 2
	  3.4 q ∧ r ∧–I, 3.2, 3.3
6.8 Prove (p → q) ∧ (q → r) → (p → r)
	(p → q) ∧ (q → r) → (p → r) by Deduction
	1 p → q ∧–E, Assumption
	2 q → r ∧–E, Assumption
	3 p → r by Deduction
	  3.1 p Assumption 3
	  3.2 q →–E, 3.1, 1
	  3.3 r →–E, 3.2, 2
6.9 Prove (q → r) → ((p → q) → (p → r))
	(q → r) → ((p → q) → (p → r)) by Deduction
	1 q → r Assumption
	2 (p → q) → (p → r) by Deduction
	  2.1 p → q Assumption 2
	  2.2 p → r by Deduction
		  2.2.1 p Assumption 2.2
		  2.2.2 q →–E, 2.2.1, 2.1
		  2.2.3 r →–E, 2.2.2, 1
6.10 Prove (p → (q → r)) → ((p → q) → (p → r))
	(p → (q → r)) → ((p → q) → (p → r)) by Deduction
	1 p → (q → r) Assumption
	2 (p → q) → (p → r) by Deduction
	  2.1 p → q Assumption 2
	  2.2 p → r by Deduction
		  2.2.1 p Assumption 2.2
		  2.2.2 q → r →–E, 2.2.1, 1
		  2.2.3 q →–E, 2.2.1, 2.1
		  2.2.4 r →–E, 2.2.3, 2.2.2
6.11 Prove (p → ¬p) → ¬p
	 (p → ¬p) → ¬p by Deduction
	 1 p → ¬p Assumption
	 2 ¬p by Contradiction
	   2.1 p Assumption
	   2.2 ¬p →–E, 2.1, 1
	   2.3 p ∧ ¬p ∧–I, 2.1, 2.2
6.12 Prove (p ∧ ¬q) → ¬(p → q)
	 (p ∧ ¬q) → ¬(p → q) by Deduction
	 1 p ∧–E, Assumption
	 2 ¬q ∧–E, Assumption
	 3 ¬(p → q) by Contradiction
	   3.1 p → q Assumption
	   3.2 q →–E, 1, 3.1
	   3.3 q ∧ ¬q ∧–I, 3.2, 2
6.13 Prove (p → q) ∧ (p → ¬q) → ¬p
	 (p → q) ∧ (p → ¬q) → ¬p by Deduction
	 1 (p → q) ∧–E, Assumption
	 2 (p → ¬q) ∧–E, Assumption
	 3 ¬p by Contradiction
	   3.1 p Assumption
	   3.2 q →–E, 3.1, 1
	   3.3 ¬q →–E, 3.1, 2
	   3.4 q ∧ ¬q ∧–I, 3.2, 3.3
6.14 Prove of p ∨ ¬p.
	 p ∨ ¬p by Contradiction
	 1 ¬(p ∨ ¬p) Assumption
	 2 ¬p ∧ ¬¬p De Morgan
6.15 Prove ¬(p ∧ ¬p).
	 1 ¬p ∨ ¬¬p De Morgan
	 2 ¬p ∨ ¬¬p Substitution, (6.14)
6.16 Prove p → q ≡ ¬q → ¬p
	 p → q ≡ ¬q → ¬p by Mutual implication
	 1 (p → q) → (¬q → ¬p) by Deduction
	   1.1 p → q Assumption 1
	   1.2 ¬q → ¬p by Deduction
		   1.2.1 ¬q Assumption 1.2
		   1.2.4 ¬p by Contradiction
				 1.2.4.1 p Assumption
				 1.2.4.2 q →–E, 1.2.4.1, 1.1
				 1.2.4.3 q ∧ ¬q, ∧–I, 1.2.4.2, 1.2.1
	 2 (¬q → ¬p) → (p → q) by Deduction
	   2.1 ¬q → ¬p Assumption 2
	   2.2 p → q by Deduction
		   2.2.1 p Assumption 2.2
		   2.2.4 q by Contradiction
				 2.2.4.1 ¬q Assumption
				 2.2.4.2 ¬p →–E, 2.2.4.1, 2.1
				 2.2.4.3 p ∧ ¬p, ∧–I, 2.2.1, 2.2.4.2
6.17 Prove p → q ≡ ¬p ∨ q
	 p → q ≡ ¬p ∨ q by Mutual implication
	 1 (p → q) → (¬p ∨ q) by Deduction
	   1.1 p → q Assumption 1
	   1.2 ¬p ∨ q by Contradiction
		   1.2.1 ¬(¬p ∨ q) Assumption
		   1.2.2 p ∧ ¬q De Morgan 1.2.1
		   1.2.3 p ∧–E, 1.2.2
		   1.2.4 ¬q ∧–E, 1.2.2
		   1.2.5 q →–E, 1.2.3, 1.1
		   1.2.6 q ∧ ¬q ∧–I, 1.2.5, 1.2.4
	 2 (¬p ∨ q) → (p → q) by Deduction
	   2.1 ¬p ∨ q Assumption 2
	   2.2 p → q by Deduction
		   2.2.1 p Assumption 2.2
		   2.2.2 q by Contradiction
				 2.2.2.1 ¬q Assumption
				 2.2.2.2 p ∧ ¬q ∧–I, 2.2.1, 2.2.2.1
				 2.2.2.3 ¬(p ∧ ¬q) De Morgan 2.1
				 2.2.2.4 (p ∧ ¬q) ∧ ¬(p ∧ ¬q) ∧–I, 2.2.2.2, 2.2.2.3
7 Formal logical systems
Proof theory'
A formal logical system, or logic, is a set of rules defined in terms of
1. a set of symbols,
2. a set of formulas constructed from the symbols,
3. a set of distinguished formulas called axioms, and
4. a set of inference rules.
7.1 Definition. A logic is consistent if at least one of its formulas is a theorem
	and at least one is not; otherwise, the logic is inconsistent.
PQ-L
Symbols: P,Q,–
Formulas: Sequences of the form a Pb Q c, where a, b, and c denote finite sequences
of zero or more dashes –.
Axioms: 0: ––P–Q––
		1: ––P–q–––
Inference Rule: a P b Q c, d P e Q f
				————————————————————
				   a d P b e Q c f
7.2 Addition-equality Interpretation. A formula a P b Q c is mapped to #a + #b = #c,
	where #x denotes the number of dashes in sequence x.
7.3 Addition-inequality Interpretation. A formula a P b Q c is mapped to true iff
	#a+#b≤#c, where #x denotes the number of dashes in sequence x.
7.4 Standard interpretation of expressions of E. For an expression P without	
	variables, let eval.'P' be the value of P, as explained in Sec 2.1. (Note that
	eval gives "meaning" to the operators and constants of E.) Let Q be any
	expression, and let s be a state that gives values to all the variables of Q.
	Define s.'Q' to be a copy of Q in which all its variables are replaced by their
	corresponding values in state s. Then function f given by f.'Q'=eval(s.'Q') is
	an interpretation for Q. 
7.5 Definition. Let S be a set of interpretations for a logic and F be a formula
	of the logic. F is satisfiable (under S) iff at least one interpretation of S
	maps F to true. F is valid (under S) iff every interpretation in S maps F to true.
7.6 Definition. A logic is sound iff every theorem is valid. A logic is complete iff
	every valid formula is a theorem.
7.7 Decision Procedure for logic E. Compute the interpretation of a formula F in
	every possible state of the state space defined by the boolean variable in F.
	F is a theorem of E iff it is mapped to true in every state.
7.8 Theorem. There exist two irrational numbers b and c such that bᶜ is rational.
7.9 P ⊢ false
	—————————
	   ¬P
7.10 ¬P ⊢ false
	——————————
		P
7.11 ⊢ (p ∨ (p → false) → false) → false
	 1 p ∨ (p → false) → false ⊢ false
	   1.1 p ⊢ false
		   1.1.1 p ∨ (p → false)         ∨–Il
		   1.1.2 false                   →–E, 1.1.1, pr 1.a
	   1.2 p → false                     →–I, 1.1
	   1.3 p ∨ (p → false)               ∨–Ir, 1.2
	   1.4 false                         →–E, 1.3, pr 1.a
	 2 (p ∨ (p → false) → false) → false →–I, 1
7.1 Give a finite set of axioms that can be added to PQ-L to make it sound and
	complete under Addition-Equality Interpretation(7.2).
	Answer:
	PQ
	–PQ–
	P–Q–
7.2 Consider the logic defined as follows, where a, b, and c denote finite
	sequences of zero or more o's.
	Symbols: M, I, o
	Formulas: Sequences of the form a M b I c
	Axiom: ooMooIoooo
	Inference Rule R1:   a M b I c
					   —————————————
					   a a M b I c c
	Inference Rule R2: a M b b I c c
					   —————————————
					   a a M b I c c
	(a) Give 5 formulas of this logic.
		Answer:
		1. MI
		2. oMI
		3. MoI
		4. MIo
		5. oMoI
	(b) State and prove five theorems of the logic.
		Answer:
		1. Prove: ooMooIoooo
			1. ooMooIoooo Axiom
		2. Prove: ooooMooIoooooooo
			1. ooMooIoooo Axiom
			2. ooooMooIoooooooo R1, 1
		3. Prove: ooooMoIoooo
			1. ooMooIoooo Axiom
			2. ooooMoIoooo R2, 1
		4. Prove: ooooooooMoIoooooooo
			1. ooMooIoooo Axiom
			2. ooooMooIoooooooo R1, 1
			3. ooooooooMoIoooooooo R2, 2
		5. Prove: ooooooooMoIoooooooo 
			1. ooMooIoooo Axiom
			2. ooooMoIoooo R2, 1
			3. ooooooooMoIoooooooo R1, 2
	(c) Give an interpretation of the logic that makes multiplication of
		integers a model.
		Answer:
			Multiplication-equality Interpretation. A formula a M b I c is mapped
			to true iff #a × #b = #c, where #x denotes the number of o in sequence x.
	(d) Give a formula that is true according to your interpretation but is not
		a theorem. Argue (informally) why it cannot be a theorem.
		Answer:
			MI, oMI, MoI
			Above formulas are true to my interpretation(0 × 0 = 0, 1 × 0 = 0, 0 × 1 = 0),
			but these are not theorems. Since theorems can be proved using R1 and R2 with Axiom.
7.3 Two possible definitions for soundness of an inference rule are:
	Theorem-Soundness.
		An inference rule is considered sound if a formula derived using it is
		valid whenever the premises used in the inference are theorems.
	Model-Soundness.
		An inference rule is considered sound if a formula derived using it is
		valid whenever the premises used in the inference are valid.
	What are the advantages/disadvantages of considering axiomatizations in which
	all inference rules satisfy Theorem-Soundness versus Model-Soundness?
	Answer:
		Advantages:
			Theorem-Soundness: It allows us axiomatizes arithmetic since it's
							   sound and not complete.
			Model-Soundness: Every valid formula can be proved and the theorems
							 are true statements about the domain of discourse.
		Disadvantages:
			Theorem-Soundness: Not every valid formula can be proved.
			Model-Soundness: Godel's incompleteness theorem, which states that
							 no formal logical system that axiomatizes arithmetic
							 can be both sound and complete.
7.4 Recall from Exercise 3.88 (page 67) that a formula is in conjunctive normal
	form if it is the conjunction of propositional logic formulas, each of which
	is a disjunction of boolean variables. For such a formula P, denote by #P the
	minimum number of boolean variables that, if all were false, would cause P to
	be false. Describe a procedure to calculate #P for P a formula in conjunctive
	normal form in which no variable is negated.
	(Solving this problem turns out to be useful in determining the fault-tolerance
	of a distributed system. Each variable denotes whether a give processor is
	faulty, and the formula is true iff the system has sufficient resources to work.)
	Answer:
	A formula P is mapped to #P, where #P denotes the minimum number of boolean variables
	in a single conjunct of P sequence.
7.5 This exercise concerns a new calculus, the 01¤–calculus. The symbols of the
	calculus are
	• Variables x, y, z, ….
	• The three constant symbols 0, 1, and ¤.
	• The binary infix predicate symbol =.
	• Parentheses.
	Formulas of the calculus have the form 𝛼=𝛽, where 𝛼 and 𝛽 are sequences of one
	or more constants, possibly with balanced parentheses as usual to indicate aggregation.
	Example of expressions are
		0¤=1 and (¤x)(¤¤)=10.
	By definition, if parentheses are omitted, left association is assumed, so that 0100
	is shorthand for ((01)0)0.
	The 01¤–calculus has inference rules Leibniz, Substitution, and Transitivity
	of equality =. Note that symmetry and reflexivity of = are not axioms, so be
	extremely careful in applying inference rules.
	There are four axioms:
		Left zero: 0¤=1
		Zero: x0¤=x1
		Left one:1¤=10
		One: x1¤=x¤0
	A theorem of the calculus is either an axiom or an expression X=Y such that
	X can be transformed into Y using the inference rules. As an example, we prove
	that 011¤0=1000 is a theorem.
		011¤0
	 =   <Axiom One, with x:=01>
		01¤00
	 =   <Axiom One, with x:=0>
		0¤000
	 =   <Axiom Left zero>
		1000
	We now give meaning to the 01¤–calculus by defining interpretations for it.
	A state assigns to each variable a natural number or ¤. For example, the state
	{(x, 19), (y, ¤)} assigns 19 to x and ¤ to y. Expressions are evaluated in a
	state by first replacing each variable by its value in the state and then applying
	the following rules. (In the rules, x ≻ y means that x evaluates to y).
		mn ≻ (2•m + n) (for integers m and n)
		m¤ ≻ (m + 1)   (for integer m)
		¤n ≻ (2 + n)    (for integer n)
		¤¤ ≻ 2
		(x = x) ≻ true  (for x an integer)
		(x = y) ≻ false (for x and y different integers)
	Perform the following exercises.
	(a) Prove 0¤¤=10.
		Answer:
			0¤¤
		 =   <Axiom Left zero>
			1¤
		 =   <Axiom Left one>
			10
	(b) Prove 0¤¤¤=11.
			0¤¤¤
		 =   <Axiom Left zero>
			1¤¤
		 =   <Axiom Left one>
			10¤
		 =   <Axiom Zero, with x:=1>
			11
	(c) Prove 0¤¤¤¤=100.
			0¤¤¤¤
		 =   <Axiom Left zero>
			1¤¤¤
		 =   <Axiom Left one>
			10¤¤
		 =   <Axiom Zero, with x:=1>
			11¤
		 =   <Axiom One, with x:=1>
			1¤0
		 =   <Axiom Left zero>
			100
	(d) Prove 0¤¤¤¤¤=101.
			0¤¤¤¤¤
		 =   <Axiom Left zero>
			1¤¤¤¤
		 =   <Axiom Left one>
			10¤¤¤
		 =   <Axiom Zero, with x:=1>
			11¤¤
		 =   <Axiom One, with x:=1>
			1¤0¤
		 =   <Axiom Zero, with x:=1¤>
			1¤1
		 =   <Axiom Left one>
			101
	(e) Why does the following metatheorem concerning the 01¤–calculus hold:
		Every theorem of the 01¤calculus contains a ¤.
		Answer:
		A theorem of the calculus is either an axiom or an expression X=Y such that
	X can be transformed into Y using the inference rules.
	(f) Does it follow from the previous question that the 01¤–calculus is consistent?
		Yes, because we can create a formula that has ¤ but not a theorem.
	(g) Evaluate the expression 1011 in the state {(x, 19), (y, ¤)}. In doing these
		and other evaluations, fully parenthesize the expression in order to be able
		to distinguish characters 0 and 1 from integers. Thus, evaluate the expression
		(((1)0)1)1.
		Answer:
			(((1)0)1)1
		 =
			((2•1 + 0)1)1
		 =
			((2)1)1
		 =
			((2•2 + 1)1
		 =
			(5)1
		 =
			2•5 + 1
		 =
			11
	(h) Evaluate the expression 1011¤ in the state {(x, 19), (y, ¤)}.
		Answer:
			((((1)0)1)1)¤
		 =
			(((2•1 + 0)1)1)¤
		 =
			(((2)1)1)¤
		 =
			((2•2 + 1)1)¤
		 =
			((5)1)¤
		 =
			(2•5 + 1)¤
		 =
			(11)¤
		 =
			11 + 1
		 =
			12
	(i) Evaluate the expression 1011¤¤ in the state {(x, 19), (y, ¤)}.
		Answer:
			(((((1)0)1)1)¤)¤
		 =
			((((2•1 + 0)1)1)¤)¤
		 =
			((((2)1)1)¤)¤
		 =
			(((2•2 + 1)1)¤)¤
		 =
			(((5)1)¤)¤
		 =
			((2•5 + 1)¤)¤
		 =
			((11)¤)¤
		 =
			(11 + 1)¤
		 =
			(12)¤
		 =
			12 + 1
		 =
			13
	(j) Evaluate the expression 1x0y¤ in the state {(x,19), (y, ¤)}.
		Answer:
			((((1)x)0)y)¤
		 =
			((((1)19)0)¤)¤
		 =
			((((1)19)0)¤)¤
		 =
			(((2•1 + 19)0)¤)¤
		 =
			(((21)0)¤)¤
		 =
			((2•21 + 0)¤)¤
		 =
			((42)¤)¤
		 =
			((42 + 1)¤
		 =
			(43)¤
		 =
			43 + 1
		 =
			44
	(k) Evaluate the expression 0(¤y)=10 in the state {(x, 19), (y, ¤)}.
		Answer:
			0(¤y)=10
		 =
			0(¤¤)=10
		 =
			0(2)=10
		 =
			2•0 + 2 = 10
		 =
			2 = 10
		 =
			false
	(l) Find a model and a counterexample for the expression x¤0=10.
		Answer:
		Model:
			(x, 0)
		Counterexample:
			(x, 1)
	(m) Show that the 01¤–calculus with the interpretation given above is sound
		by checking that all four axioms are valid and that all three inference rules
		preserve validity.
		Answer:
		Axiom Left zero:
			(((0)¤)=1
		 =
			(0 + 1)=1
		 =
			1=1
		 =
			true
		Axiom Zero:
			(((x)0)¤)=(x)1
		 =
			((2•x + 0)¤)=(x)1
		 =
			((2•x)¤)=(x)1
		 =
			2•x + 1=(x)1
		 =
			2•x + 1 = 2•x + 1
		Axiom Left one:
			(1)¤=(1)0
		 =
			1 + 1 = (1)0
		 =
			2 = (1)0
		 =
			2 = 2•1 + 0
		 =
			2 = 2
		 =
			true
		Axiom One:
			((x)1)¤=((x)¤)0
		 =
			(2•x + 1)¤=((x)¤)0
		 =
			2•x + 1 + 1=((x)¤)0
		 =
			2•x + 2=((x)¤)0
		 =
			2•x + 2=(x + 1)0
		 =
			2•x + 2=2•(x + 1)+0
		 =
			2•x + 2=2•x + 2
		 =
			true
		All four axioms evals to true in any interpretation, therefore they are valid.
		Because all formulas in 01¤–calculus are in the form of A=B, 
		based on the two rules in the interpretation:
			(x = x) ≻ true  (for x an integer)
			(x = y) ≻ false (for x and y different integers)
		We can conclude that in 01¤–calculus, the interpretation of "=" is just
		the equality in the metalanguage(the domain of integers). 
		Therefore, all three inference rules are valid.
	(n) Show that the 01¤–calculus is incomplete by finding a valid expression that
		is not a theorem.
		Answer:
		0=0
		Because the conclusions of these three inference rules, have LHS and
		RHS lengths that are not less than the premise's LHS and RHS, it implies
		that the LHS and RHS lengths of any theorem will definitely not be less
		than the LHS and RHS of the axioms(premise).
		Therefore, 0=0 is not a theorem.
	(o) Show that the expression x1=¤¤ is unsatisfiable.
		Answer:
			x1=¤¤
		 =
			2•x + 1 = 2
		Since 2•x+1=2 has no integer solution and ¤1=¤¤=3, x1=¤¤ is unsatisfiable.
8 On types
Some basic types
Name Symbol Type(set of values)
integer ℤ integers: …, -3, -2, -1, 0, 1, 2, 3, …
nat ℕ integers: 0, 1, 2, …
positive ℤ⁺ positive integers: 1, 2, 3, …
negative ℤ⁻ negative integers: -1, -2, -3, …
rational ℚ rational numbers i/j for i,j integers, j ≠ 0
reals ℝ⁺ positive real numbers
bool 𝔹 booleans: true, false
8.1 f:t1×…×tn → r
8.2 i ∈ E → -i ≤ 0
If E has type t, i.e. E:t, then E ∈ t evaluates to true in all states in which
E is well defined.
8.3 In a textual substitution E[x:=F], x and F must have the same type.
8.4 Equality b=c is defined on ly if b and c have the same types. That is,
	equality = has type t × t → 𝔹, for any type t.
8.5 ∑[i=1…n]e
	Formula above stands for e[i:=1] + e[i:=2] +…+ e[i:=n] —in words, for the
	sum of the values e[i:=v] for integers v in the range i…n. Here is an example
	∑[i=1…3]i² = 1² + 2² + 3²
	(∑i ∣ 1 ≤ i ≤ n:e) or (+i ∣ 1 ≤ i ≤ n:e)
8.6 (★x:t1, y:t2 ∣ R:P)
8.7 (∧i ∣ : x•i=0)
8.8 i > 0 ∨ (∧i ∣ 0≤i : x•i=0)
8.9 Definition. The occurrence of i in the expression i is free.
	Suppose an occurrence of i in expression E is free. Then that
	same occurrence of i is free in (E), in function application f(…,E,…),
	and in (★x ∣ E:F) and (★x ∣ F:E) provided i is not one of the dummies in list x.
	Define occurs('v', 'e') to mean that at least one variable in the list v of
	variables occurs free in at least one expression in expression list e.
8.10 Definition. Let an occurrence of i be free in an expression E. That occurrence
	of i is bound (to dummy i) in the expression (★x ∣ E:F) or (★x ∣ F:E) if i is
	one of the dummies in list x.
	Suppose an occurrence of i is bound in expression E. Then it is also bound
	(to the same dummy) in (E), f(…,E,…), (★x ∣ E:F) and (★x ∣ F:E).
	Example: i + j + (∑i ∣ 1≤i≤10 : b[i]ʲ) +
			 (∑i ∣ 1≤i≤10 : (∑j ∣ 1≤j≤10 : c[i,j]))
	1. The leftmost occurrence of i is free in this expression.
	2. The leftmost occurrence of j and the occurrence of j as an exponent
	   in the first summation are both free.
	3. All other occurrences of i and j are bound. There are two different
	   dummies i.
8.11 Provided ¬occurs('y', 'x, F'),
		(★y ∣ R:P)[x:=F] = (★y ∣ R[x:=F] : P[x:=F])
		Examples:
		(+x ∣ 1≤x≤2 : y)[y:=y+z]
			= (+x ∣ 1≤x≤2 : y+z)
		(+i ∣ 0≤i<n : b[i]=n)[n:=m]
			= (+i ∣ 0≤i<m : b[i]=m)
		(+y ∣ 0≤y<n : b[y]=n)[n:=y]
			= (+j ∣ 0≤j<y : b[j]=y)
		(+y ∣ 0≤y<n : b[y]=n)[y:=m]
			= (+j ∣ 0≤j<n : b[j]=n)
8.12 Leibniz:
		P = Q
	———————————————————————————————————————
	(★x ∣ E[z:=P] : S) = (★x ∣ E[z:=Q] : S)
		   R → P = Q
	———————————————————————————————————————
	(★x ∣ R : E[z:=P]) = (★x ∣ R : E[z:=Q])
8.13 Axiom, Empty range: (★x ∣ false : P) = u (the identity of ★)
8.14 Axiom, One-point rule:
	 Provided ¬occurs('x', 'E'),
	 (★x ∣ x=E : P) = P[x:=E]
		Example:
		(+x ∣ x=3 : x²) = 3²
8.15 Axiom, Distributivity: Provided each quantification is defined,
	 (★x ∣ R : P) ★ (★x ∣ R : Q) = (★x ∣ R : P★Q)
8.16 Axiom, Range split: Provided R ∧ S ≡ false and each quantification is defined,
	 (★x ∣ R∨S : P) = (★x ∣ R : P) ★ (★x ∣ S : P)
	 Example:
		 (+x ∣ R∨S : P) = (+x ∣ R : P) + (+x ∣ S : P)
		 R = 1,2,3,4
		 S = 8,9
		 R∨S=1,2,3,4,8,9
		 (+i ∣ R∨S : i*2) = 1*2 + 2*2 + 3*2 + 4*2 + 8*2 + 9*2
					 = (1 + 2 + 3 + 4 + 8 + 9) * 2
					 = 27 * 2
					 = 54
		 (+i ∣ R : i*2) = 1*2 + 2*2 + 3*2 + 4*2
				   = (1 + 2 + 3 + 4) * 2
				   = ((1+4) * 4) / 2 * 2
				   = 20
		 (+i ∣ S : i*2) = 8*2 + 9*2
				   = 17 * 2
				   = 34
		 (+i ∣ R : i*2) + (+i ∣ S : i*2) = 20 + 34
								= 54
8.17 Axiom, Range split: Provided each quantification is defined,
	 (★x ∣ R∨S : P) ★ (★x ∣ R∧S : P) = (★x ∣ R : P) ★ (★x ∣ S : P)
	 Example:
		 (+x ∣ R∨S : P) + (+x ∣ R∧S : P) = (+x ∣ R : P) + (+x ∣ S : P)
		 R = 1,2,3,4
		 S = 4,5
		 R∨S=1,2,3,4,5
		 R∧S=4
		 (+x ∣ R∨S : P) = 1 + 2 + 3 + 4 + 5
						= 15
		 (+x ∣ R∧S : P) = 4
		 (+x ∣ R : P) = 1 + 2 + 3 + 4
					  = 10
		 (+x ∣ S : P) = 4 + 5
					  = 9
		 (+x ∣ R∨S : P) + (+x ∣ R∧S : P) = 19
			 (+x ∣ R : P) + (+x ∣ S : P) = 19
8.18 Axiom, Range split for idempotent ★: Provided each quantification is defined,
	 (★x ∣ R∨S : P) = (★x ∣ R : P) ★ (★x ∣ S : P)
	 Example:
		 (∧x ∣ R∨S : P) = (∧x ∣ R : P) ∧ (∧x ∣ S : P)
		 R = true,false
		 S = true
		 R∨S = true,false
		 (∧x ∣ R∨S : P) = true ∧ false
						= false
		 (∧x ∣ R : P) = true ∧ false
					  = false
		 (∧x ∣ S : P) = true
		 (∧x ∣ R : P) ∧ (∧x ∣ S : P) = false ∧ true
									 = false
8.19 Axiom, Interchange of dummies: Provided each quantification is defined,
	 ¬occurs('y', 'R') and ¬occurs('x', 'Q')
		(★x ∣ R : (★y ∣ Q : P)) = (★y ∣ Q :(★x ∣ R : P))
	 Example:
		 (+x ∣ R : (+y ∣ Q : P)) = (+y ∣ Q :(★x ∣ R : P))
		 R = 1,2,3,4
		 Q = 4,5
		 (+x ∣ R : (+y ∣ Q : P)) = ((1 + 4) + (1 + 5)) +
								   ((2 + 4) + (2 + 5)) +
								   ((3 + 4) + (3 + 5)) +
								   ((4 + 4) + (4 + 5))
								 = (((1 + 4) × 4) ÷ 2) × 2 + (4 + 5) × 4
								 = 20 + 36
								 = 56
		 (+y ∣ Q :(★x ∣ R : P)) = ((4 + 1) + (4 + 2) + (4 + 3) + (4 + 4)) +
								  ((5 + 1) + (5 + 2) + (5 + 3) + (5 + 4))
								= ((4 + 5) × 4) + (((1 + 4) × 4) ÷ 2) × 2
								= 36 + 20
								= 56
8.20 Axiom, Nesting: Provided ¬occurs('y', 'R'),
	 (★x,y ∣ R∧Q : P) = (★x ∣ R : (★y ∣ Q : P))
	 Example:
		 (+x,y ∣ R∧Q : P) = (+x ∣ R : (+y ∣ Q : P))
		 R = 1,2,3,4
		 Q = 4,5
		 (+x,y ∣ R∧Q : P) = ((1 + 4) + (1 + 5)) + ((2 + 4) + (2 + 5)) +
							((3 + 4) + (3 + 5)) + ((4 + 4) + (4 + 5))
						  = (1 + 2 + 3 + 4) × 2 + (4 + 5) × 4
						  = 20 + 36
						  = 56
		 (+x ∣ R : (+y ∣ Q : P)) = ((1 + 4) + (1 + 5)) +
								   ((2 + 4) + (2 + 5)) +
								   ((3 + 4) + (3 + 5)) +
								   ((4 + 4) + (4 + 5))
								 = (((1 + 4) × 4) ÷ 2) × 2 + (4 + 5) × 4
								 = 20 + 36
								 = 56
	 Note:
		¬occurs('y', 'R')
		In the case of occurrence:
			Since it occurs and is not free, it must be bound.
			Therefore, it can be renamed(8.21). (This is used to prevent
			quantification within boolean expressions.
			Usually, R is a simple boolean expression with no quantification.)
			Assume that the quantification range for x is R, and the
			quantification range for y is Y.
		In the case of non-occurrence:
			Assume that the quantification range for x is R, and
			the quantification range for y is Y.
8.21 Axiom, Dummy renaming: Provided ¬occurs('y', 'R,P')
	 (★x ∣ R : P) = (★y ∣ R[x:=y] : P[x:=y])
	 Example:
	 (+x ∣ R : P) = (+y ∣ R[x:=y] : P[x:=y])
	 R = 1,2,3,4
	 (+x ∣ R : P) = 1 + 2 + 3 + 4
	 (+y ∣ R[x:=y] : P[x:=y]) = 1 + 2 + 3 + 4
8.22 Change of dummy: Provided ¬occurs('y', 'R,P') and f has an inverse,
	 (★x ∣ R : P) = (★y ∣ R[x:=f.y] : P[x:=f.y])
		(★y ∣ R[x:=f.y] : P[x:=f.y])
	 =   <One-point rule(8.14)
		  —Quantification over x has to be introduced. The One-point rule is the
		  only rule that can be applied at first.>
		(★y ∣ R[x:=f.y] : (★x ∣ x=f.y : P))
	 =   <Nesting(8.20) —Moving dummy x to the outside gets us closer to the final
		  form.>
		(★x,y ∣ R[x:=f.y]∧x=f.y : P)
	 =   <Substitution(3.84a) —R[x:=f.y] must be removed at some point.
		  This substitution makes it possible.>
		(★x,y ∣ R[x:=x]∧x=f.y : P)
	 =   <R[x:=x]≡R; Nesting(8.20), ¬occurs('y', 'R')
		  —Now we can get a quantification in x alone.>
		(★x ∣ R : (★y ∣ x=f.y : P))
	 =   <x = f.y ≡ y = f^-1.x —This step prepares for the elimination of y using
		  the One-point rule.>
		(★x ∣ R : (★y ∣ y=f^-1.x : P))
	 =   <One-point rule(8.14)>
		(★x ∣ R : P[y:=f^-1.x])
	 =   <Textual substitution —¬occurs('y', 'P')>
		(★X ∣ R : P)
8.23 Theorem Split off term. For n:ℕ and dummies i:ℕ,
	 (★i ∣ 0≤i<n+1 : P) = (★i ∣ 0≤i<n : P) ★ P[i:=n]
	 (★i ∣ 0≤i<n+1 : P) = P[i:=n] ★ (★i ∣ 0<i<n+1 : P)
	 Proof.
		(★i ∣ 0≤i<n+1 : P)
	 =   <0≤i<n+1 ≡ 0≤i<n∨i=n>
		(★i ∣ 0≤i<n∨i=n : P)
	 =   <Range split(8.16) —0≤i<n ∧ i=n ≡ false>
		(★i ∣ 0≤i<n : P) ★ (★i ∣ i=n : P)
	 =   <One-point rule(8.14)>
		(★i ∣ 0≤i<n : P) ★ P[i:=n]
8.24 b≤c≤d → (b≤i<d ≡ b≤i<c ∨ c≤i<d)
8.25 (∑i,j ∣ 0≤i≤j<n+1 : c[i,j])
Proof.
	0≤i≤j<n+1
 =   <Remove abbreviation>
	0≤i≤j ∧ j<n+1
 =   <j<n+1 ≡ j<n∨j=n>
	0≤i≤j ∧ (j<n∨j=n)
 =   <Distributivity of ∧ over ∨>
	(0≤i≤j ∧ j<n) ∨ (0≤i≤j ∧ j=n)
 =   <Reintroduce abbreviation>
	(0≤i≤j<n) ∨ (0≤i≤j ∧ j=n)
	(∑i,j ∣ 0≤i≤j<n+1 : c[i,j])
 =   <Above proof>
	(∑i,j ∣ (0≤i≤j<n)∨(0≤i≤j ∧ j=n) : c[i,j])
 =   <Range split(8.16) —0≤i≤j<n ∧ 0≤i≤j∧j=n ≡ false>
	(∑i,j ∣ (0≤i≤j<n) : c[i,j]) + (∑i,j ∣ (0≤i≤j∧j=n) : c[i,j])
 =   <Nesting(8.20), with x,y,R,Q,P:=j,i,j=n,0≤i≤j,c[i,j]>
	(∑i,j ∣ (0≤i≤j<n) : c[i,j]) + (∑j ∣ j=n : (∑i ∣ 0≤i≤j : c[i,j]))
 =   <One-point rule(8.14), with x,E,P:=j,n,(∑i∣0≤i≤j:c[i,j])>
	(∑i,j ∣ (0≤i≤j<n) : c[i,j]) + (∑i ∣ 0≤i≤n : c[i,n])
8.1 Given are functions a, b, c, d, and e with types as follows.
	 a : A → B
	 b : B → C
	 c : C → A
	 d : A × C → D
	 e : B × B → E
	 State whether each expression below is type correct. If not, explain why.
	 Assume u:A, w:B, x:C, y:D, and z:E.
		(a) e(a.u, w)
			Answer:
				It is type correct.
		(b) b.x
			Answer:
				It is not type correct because the actual argument x of 
				function b has type C. This does not match with b's accepted 
				formal parameter type B.
		(c) e(a(c.x), a.u)
			Answer:
				It is type correct.
		(d) a(c(b(a.y)))
			Answer:
				It is not type correct because the actual argument y of 
				function a has type D. This does not match with a's accepted 
				formal parameter 		type A.
		(e) d(c.x, c.x)
			Answer:
				It is not type correct because the second actual argument c.x of 
				function d has type A. This does not match with d's second accepted 
				formal parameter type C
8.2 Consider the expression e ∈ t where t has type set(ℤ), i.e. set of integers.
	Give a reasonable type for function ∈ and for function application e ∈ t.
	 Answer:
		∈ : set × set → 𝔹
		e ∈ t : 𝔹
8.3 Expand the following textual substitutions. If necessary, change the dummy,
	according to Dummy Renaming(8.21).
		(a) (★x ∣ 0≤x+r<n : x+v)[v:=3]
			Answer:
				(★x ∣ 0≤x+r<n : x+3) 
		(b) (★x ∣ 0≤x+r<n : x+v)[x:=3]
			Answer:
				(★y ∣ 0≤y+r<n : y+v)
		(c) (★x ∣ 0≤x+r<n : x+v)[n:=n+x]
			Answer:
				(★x ∣ 0≤x+r<n+x : x+v)
		(d) (★x ∣ 0≤x<r : (★y ∣ 0≤y : x+y+n))[n:=x+y]
			Answer:
				(★x ∣ 0≤x<r : (★z ∣ 0≤z : x+z+x+y))
		(e) (★x ∣ 0≤x<r : (★y ∣ 0≤y: x+y+n))[r:=y]
			Answer:
				(★x ∣ 0≤x<y : (★z ∣ 0≤z: x+z+n))
8.4 Give a definition of E[x:=e] for all expressions E, including quantifications.
	The definition should be in terms of the different kinds of expression E, just
	as the notions of free and bound were defined. Treat expressions that are constants,
	variables, parenthesized expressions, unary operations, binary operations,
	function appilcations, and quantificiation.
	Answer:
		Definition.
			constants:
				Nil!
			variables:
				x[x:=e]=e
				y[x:=e]=y
			parenthesized expressions:
				(P)[x:=e]=(P[x:=e])
			unary operations:
				★P[x:=e]=★(P[x:=e])
			binary operations:
				(P★Q)[x:=e]=P[x:=e]★Q[x:=e]
			quantificiation:
				(★x∣R:P)[x:=e]=Nil!
				(★y∣R:P)[x:=e]
				case occurs('y', 'e') of
				true:
					(★y∣R:P)[x:=e]=(★z|R[y:=z]:P[y:=z])[x:=e]
				false:
					(★y∣R:P)[x:=e]=(★y|R[x:=e]:P[x:=e])
			function applications:
				(P.Q)[x:=e]=P[x:=e].Q[x:=e]
8.5 Prove the following theorems. Provided 0≤n,
	 (a) (∑i ∣ 0≤i<n+1 : b[i]) = b[0] + (∑i ∣ 1≤i<n+1 : b[i])
		 Answer:
			Proof.
				(∑i ∣ 0≤i<n+1 : b[i])
			 =   <0≤i<n+1 ≡ i=0∨0<i<n+1>
				(∑i ∣ i=0∨0<i<n+1 : b[i])
			 =   <Range split(8.16)>
				(∑i ∣ i=0 : b[i]) + (∑i ∣ 0<i<n+1 : b[i])
			 =   <One-point rule(8.14)>
				b[0] + (∑i ∣ 0<i<n+1 : b[i])
	 (b) (∑i ∣ 0≤i≤n : b[i]) = (∑i ∣ 0≤i<n : b[i]) + b[n]
		 Answer:
			Proof.
				(∑i ∣ 0≤i≤n : b[i])
			 =   <0≤i≤n ≡ 0≤i<n∨i=n>
				(∑i ∣ 0≤i<n∨i=n : b[i])
			 =   <Range split(8.16)>
				(∑i ∣ 0≤i<n : b[i]) + (∑i ∣ i=n : b[i])
			 =   <One-point rule(8.14)>
				(∑i ∣ 0≤i<n : b[i]) + b[n]
	 (c) (∑i ∣ 0≤i≤n : b[i]) = b[0] + (∑i ∣ 1≤i≤n : b[i])
		 Answer:
			Proof.
				(∑i ∣ 0≤i≤n : b[i])
			 =   <0≤i≤n ≡ i=0∨1≤i≤n>
				(∑i ∣ i=0∨1≤i≤n : b[i])
			 =   <Range split(8.16)>
				(∑i ∣ i=0 : b[i]) + (∑i ∣ 1≤i≤n : b[i])
			 =   <One-point rule(8.14)>
				b[0] + (∑i ∣ 1≤i≤n : b[i])
8.6 Prove the following theorems. Provided 0≤n,
	 (a) (∨i ∣ 0≤i<n+1 : b[i]=0) ≡ (∨i ∣ 0≤i<n : b[i]=0) ∨ b[n]=0
		Answer:
			Proof.
				(∨i ∣ 0≤i<n+1 : b[i]=0)
			 =   <0≤i<n+1 ≡ 0≤i<n∨i=n>
				(∨i ∣ 0≤i<n∨i=n : b[i]=0)
			 =   <Range split(8.16)>
				(∨i ∣ 0≤i<n : b[i]=0) ∨ (∨i ∣ i=n : b[i]=0)
			 =   <One-point rule(8.14)>
				(∨i ∣ 0≤i<n : b[i]=0) ∨ b[n]=0
	 (b) (∧i ∣ 0≤i<n+1 : b[i]=0) ≡ (∧i ∣ 0≤i<n : b[i]=0) ∧ b[n]=0
		Answer:
			Proof.
				(∧i ∣ 0≤i<n+1 : b[i]=0)
			 =   <0≤i<n+1 ≡ 0≤i<n∨i=n>
				(∧i ∣ 0≤i<n∨i=n : b[i]=0)
			 =   <Range split(8.16)>
				(∧i ∣ 0≤i<n : b[i]=0) ∧ (∧i ∣ i=n : b[i]=0)
			 =   <One-point rule(8.14)>
				(∧i ∣ 0≤i<n : b[i]=0) ∧ b[n]=0
	 (c) (∨i ∣ 0≤i<n+1 : b[i]=0) ≡ b[0]=0 ∨ (∨i ∣ 0<i<n+1 : b[i]=0)
		Answer:
			Proof.
				(∨i ∣ 0≤i<n+1 : b[i]=0)
			 =   <0≤i<n+1 ≡ i=0∨0<i<n+1>
				(∨i ∣ i=0∨0<i<n+1 : b[i]=0)
			 =   <Range split(8.16)>
				(∨i ∣ i=0 : b[i]=0) ∨ (∨i ∣ 0<i<n+1 : b[i]=0)
			 =   <One-point rule(8.14)>
				b[0]=0 ∨ (∨i ∣ 0<i<n+1 : b[i]=0)
	 (d) (∧i ∣ 0≤i<n+1 : b[i]=0) ≡ b[0]=0 ∧ (∧i ∣ 0<i<n+1 : b[i]=0)
		Answer:
			Proof.
				(∧i ∣ 0≤i<n+1 : b[i]=0)
			 =   <0≤i<n+1 ≡ i=0∨0<i<n+1>
				(∧i ∣ i=0∨0<i<n+1 : b[i]=0)
			 =   <Range split(8.16)>
				(∧i ∣ i=0 : b[i]=0) ∧ (∧i ∣ 0<i<n+1 : b[i]=0)
			 =   <One-point rule(8.14)>
				b[0]=0 ∧ (∨i ∣ 0<i<n+1 : b[i]=0)
8.7 Prove the following theorems:
	 (a) (+i ∣ 0≤i≤n : i) = (+i ∣ 0≤i≤n∧even.i : i) + (+i ∣ 0≤i≤n∧odd.i : i)
		Answer:
			Proof.
				(+i ∣ 0≤i≤n : i)
			 =   <0≤i≤n ≡ 0≤i≤n∧even.i∨0≤i≤n∧odd.i>
				(+i ∣ 0≤i≤n∧even.i∨0≤i≤n∧odd.i : i)
			 =   <Range split(8.16)>
				(+i ∣ 0≤i≤n∧even.i : i) + (+i ∣ 0≤i≤n∧odd.i : i)
	 (b) (+i ∣ 0≤i≤10 : 0) = 0
		Answer:
			Proof.
				(+i ∣ 0≤i≤10 : 0)
			 =   <i=0∨i=1∨i=2∨i=3∨i=4∨i=5∨i=6∨i=7∨i=8∨i=9∨i=10>
				(+i ∣ i=0∨i=1∨i=2∨i=3∨i=4∨i=5∨i=6∨i=7∨i=8∨i=9∨i=10 : 0)
			 =   <Range split(8.16)>
				(+i ∣ i=0 : 0) + (+i ∣ i=1 : 0) + (+i ∣ i=2 : 0) +
				(+i ∣ i=3 : 0) + (+i ∣ i=4 : 0) + (+i ∣ i=5 : 0) +
				(+i ∣ i=6 : 0) + (+i ∣ i=7 : 0) + (+i ∣ i=8 : 0) +
				(+i ∣ i=9 : 0) + (+i ∣ i=10 : 0)
			 =   <One-point rule(8.14) eleven times>
				0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
			 =
				0
Universal quantification
9.1 (∀x ∣ R : P)
	For all x such that R holds, P holds.
9.2 Axiom, Trading: (∀x ∣ R : P) ≡ (∀x ∣ : R→P)
9.3 Trading: (a) (∀x ∣ R : P) ≡ (∀x ∣ : ¬R∨P)
			 (b) (∀x ∣ R : P) ≡ (∀x ∣ : R∧P≡R)
			 (c) (∀x ∣ R : P) ≡ (∀x ∣ : R∨P≡P)
9.4 Trading: (a) (∀x ∣ Q∧R : P) ≡ (∀x ∣ Q : R→P)
			 Proof.
				(∀x ∣ Q∧R : P)
			 =   <Trading(9.2)>
				(∀x ∣ : Q∧R→P)
			 =   <Currying(3.65)>
				(∀x ∣ : Q→(R→P))
			 =   <Trading(9.2)>
				(∀x ∣ Q : (R→P))
			 (b) (∀x ∣ Q∧R : P) ≡ (∀x ∣ Q : ¬R∨P)
			 (c) (∀x ∣ Q∧R : P) ≡ (∀x ∣ Q : R∧P≡R)
			 (d) (∀x ∣ Q∧R : P) ≡ (∀x ∣ Q : R∨P≡P)
9.5 Axiom, Distributivity of ∨ over ∀:
	 Provided ¬occurs('x', 'P'),
		P ∨ (∀x ∣ R : Q) ≡ (∀x ∣ R : P∨Q)
9.6 Provided ¬occurs('x', 'P'),
	(∀x ∣ R : P) ≡ P ∨ (∀x ∣ : ¬R)
9.7 Distributivity of ∧ over ∀: Provided ¬occurs('x', 'P'),
	¬(∀x ∣ : ¬R) → ((∀x ∣ R : P∧Q) ≡ P ∧ (∀x ∣ R : Q))
	R → (P∧Q ≡ P∧Q)
	 Proof.
		(∀x ∣ R : P∧Q)
	 =   <Distributivity of ∀ over ∧(8.15)>
		(∀x ∣ R : P) ∧ (∀x ∣ R : Q)
	 =   <(9.6) —since ¬occurs('x', 'P')>
		(P ∨ (∀x ∣ : ¬R)) ∧ (∀x ∣ R : Q)
	 =   <Assumption ¬(∀x ∣ : ¬R), i.e. (∀x ∣ : ¬R) ≡ false>
		(P ∨ false) ∧ (∀x ∣ R : Q)
	 =   <Identity of ∨(3.30)>
		P ∧ (∀x ∣ R : Q)
9.8 (∀x ∣ R : true) ≡ true
9.9 (∀x ∣ R : P≡Q) → ((∀x ∣ R : P) ≡ (∀x ∣ R : Q))
9.10 Range weakening/strengthening:
	(∀x ∣ Q∨R : P) → (∀x ∣ Q : P)
9.11 Body weakening/strengthening:
	(∀x ∣ R : P∧Q) → (∀x ∣ R : P)
9.12 Monotonicity of ∀:
	(∀x ∣ : Q→P) → ((∀x ∣ : Q) → (∀x ∣ : P))
	Proof.
		(∀x ∣ : Q → P) → ((∀x ∣ : Q) → (∀x ∣ : P))
	 =   <Currying(3.65)>
		(∀x ∣ : Q → P) ∧ (∀x ∣ : Q) → (∀x ∣ : P)
	(∀x ∣ : Q → P) ∧ (∀x ∣ : Q) → (∀x ∣ : P) by Deduction
	1 (∀x ∣ : Q → P) Assumption
	2 (∀x ∣ : Q) Assumption
	3 P by Modus ponens
	  3.1 Q by Instantiation, 2
	  3.2 Q → P by Instantiation, 1
	4 (∀x ∣ : P) by metatheorem, 3
9.13 Instantiation: (∀x ∣ : P) → P[x:=E]
9.14 (∀i:ℤ ∣ even.i ≡ even(i²))
9.15 a+b=b+a (for all integers a,b)
	 (∀a,b:ℤ ∣ : a+b=b+a)
closed wff is also called sentence
open wff is called sentential function
9.16 Metatheorem. P is a theorem iff (∀x ∣ : P) is a theorem.
	 Proof. The proof is by mutual implication.
	 LHS → RHS. Assume P is a theorem. Then there is a proof of it that
	 transforms P to true, using Leibniz, Transitivity of equals, and
	 Substitution:
		P
	 =   <Hint 1>
		…
	…
		…
	 =   <Hint n>
		true
	 Leibniz(8.12) allows us to turn this proof into a proof of (∀x ∣ : P):
		(∀x ∣ : P)
	 =   <Hint 1>
		…
	…
		…
	 =   <Hint n>
		(∀x ∣ : true)
	 =   <(9.8), (∀x ∣ : true) ≡ true>
		true
	 RHS → LHS. Assume (∀x ∣ : P) is a theorem. Using Instantiation(9.13) with
	 x for E, we conclude that P is a theorem.
Existential quantification
	(∃x ∣ R : P)
	There exists an x in the range R such that P holds.
		A value x̂ for which (R ∧ P)[x:=x̂] is valid is called a witness for x in
		(∃x ∣ R : P).
9.17 Axiom, Generalized De Morgan: (∃x ∣ R : P) ≡ ¬(∀x ∣ R : ¬P)
		(∃x ∣ : P) ≡ ¬(∀x ∣ : ¬P)
			There exists a witness for P.
			P sometimes holds.
	 Idea behind this generalization with an example:
		(∨i ∣ 0≤i<4 : P)
	 =   <Elimiiate quantification>
		P^i_0 ∨ P^i_1 ∨ P^i_2 ∨ P^i_3
	 =   <Double negation(.12); De Morgan(3.47a)>
		¬(¬P^i_0 ∧ ¬P^i_1 ∧ ¬P^i_2 ∧ ¬P^i_3)
	 =   <Introduce quantification>
		¬(∧i ∣ 0≤i<4 : ¬P)
9.18 Generalized De Morgan: (a) ¬(∃x ∣ : ¬P) ≡ (∀x ∣ : P)
			    (b) ¬(∃x ∣ : P) ≡ (∀x ∣ : ¬P)
			    (c) (∃x ∣ : ¬P) ≡ ¬(∀x ∣ : P)
	Note: (∃x ∣ : P) ≡ ¬(∀x ∣ : ¬P)
			Disjunction of P in all states must be true.
			∵ At least one P is true.
				∴ There exists a witness for P.
			Conjunction of ¬P in all states must be false.
			∵ At least one ¬P is false.
				∴ P sometimes holds.
		¬(∃x ∣ : ¬P) ≡ (∀x ∣ : P)
			Disjunction of ¬P in all states must be false.
			∵ ¬P always false.
				∴ No witness of ¬P.
			Conjunctional of P in all states must be true.
			∵ P always true.
				∴ P always holds.
		¬(∃x ∣ : P) ≡ (∀x ∣ : ¬P)
			Disjunction of P in all states must be false.
			∵ P always false.
				∴ No witness of P.
			Conjunction of ¬P in all states must be true.
			∵ ¬P always true.
				∴ P never holds.
		(∃x ∣ : ¬P) ≡ ¬(∀x ∣ : P)
			Disjunction of ¬P in all states must be true.
			∵ At least one ¬P is true.
				∴ There exists a witness of ¬P.
			Conjunction of P in all states must be false.
			∵ At least one P is false.
				∴ P doesn't always hold.
		∴ We can't replace "P sometimes holds." by "P doesn't always hold.".
9.19 Trading: (∃x ∣ R : P) ≡ (∃x ∣ : R∧P)
	 LHS: There exists a value x in the range R for which P is true.
	 RHS: There is a value x for which both R and P is true.
		  Disjunction of R∧P in all states must be true
		  ∵ At least one R∧P is true
			∴ There exists a witness of R∧P
9.20 Trading: (∃x ∣ Q∧R : P) ≡ (∃x ∣ Q : R∧P)
9.21 Distributivity of ∧ over ∃: Provided ¬occurs('x', 'P'),
	 P ∧ (∃x ∣ R : Q) ≡ (∃x ∣ R : P∧Q)
9.22 Provided ¬occurs('x', 'P'),
	 (∃x ∣ R : P) ≡ P ∧ (∃x ∣ : R)
	 Proof.
		(∃x ∣ R : P)
	 =   <Trading(9.19)>
		(∃x ∣ : R∧P)
	 =   <Distributivity of ∧ over ∃(9.21)>
		P ∧ (∃x ∣ : R)
9.23 Distributivity of ∨ over ∃: Provided ¬occurs('x', 'P'),
	 (∃x ∣ : R) → ((∃x ∣ R : P∨Q) ≡ P ∨ (∃x ∣ R : Q))
	 Proof.
	 Assume (∃x ∣ : R)
		(∃x ∣ R : P∨Q)
	 =   <Distributivity(8.15)>
		(∃x ∣ R : P) ∨ (∃x ∣ R : Q)
	 =   <(9.22)>
		P ∧ (∃x ∣ : R) ∨ (∃x ∣ R : Q)
	 =   <Assumption>
		P ∧ true ∨ (∃x ∣ R : Q)
	 =   <Identity of ∧(3.39)>
		P ∨ (∃x ∣ R : Q)
9.24 (∃x ∣ R : false) ≡ false
9.25 Range weakening/strengthening:
	 (∃x ∣ R : P) → (∃x ∣ Q∨R : P)
9.26 Body weakening/strengthening:
	 (∃x ∣ R : P) → (∃x ∣ R : P∨Q)
9.27 Monotonicity of ∃:
	 (∀x ∣ R : Q→P) → ((∃x ∣ R : Q) → (∃x ∣ R : P))
9.28 ∃–Introduction: P[x:=E] → (∃x ∣ : P)
9.29 Interchange of quantifications
	 Provided ¬occurs('y', 'R') and ¬occurs('x', 'Q'),
	 (∃x ∣ R : (∀y ∣ Q : P)) → (∀y ∣ Q : (∃x ∣ R : P))
	 Proof.
		(∃x ∣ : (∀y ∣ : P)) → (∀y ∣ : (∃x ∣ : P))
	 =   <Implication(3.57), p→q≡p∨q≡q, to eliminate the problematic →>
		(∃x ∣ : (∀y ∣ : P)) ∨ (∀y ∣ : (∃x ∣ : P)) ≡ (∀y ∣ : (∃x ∣ : P))
	 =   <Distributivity of ∨ over ∀(9.5) —so that the LHS and RHS have
		  the same outer quantification.>
		(∀y ∣ : (∃x ∣ : P) ∨ (∃x ∣ : (∀y ∣ : P))) ≡ (∀y ∣ : (∃x ∣ : P))
	 =   <Distributivity(8.15) —so that LHS and RHS have the same two outer
		  quantifications.>
		(∀y ∣ : (∃x ∣ : (∀y ∣ : P)∨P)) ≡ (∀y ∣ : (∃x ∣ : P))
	 =   <Instantiation(9.13) says (∀y ∣ : P)→P, which by (3.57) is equivalent
		  to (∀y∣:P)∨P≡P>
		(∀y ∣ : (∃x ∣ : P)) ≡ (∀y ∣ : (∃x ∣ : P)) —Reflexivity of equality
9.30 Metatheorem Witness. Suppose ¬occurs('x̂', 'P, Q, R'). Then
	 (∃x ∣ R : P) → Q is a theorem iff (R ∧ P)[x:=x̂] → Q is a theorem.
	 Identifier x̂ is called a witness for the existential quantification.
	 Proof.
		(∃x ∣ R : P) → Q
	 =   <Trading(9.19)>
		(∃x ∣ : R∧P) → Q
	 =   <Implication(3.59); De Morgan(9.18b)>
		(∀x ∣ : ¬(R∧P)) ∨ Q
	 =   <Dummy renaming(8.21), —¬occurs('x̂', 'P, R')>
		(∀x̂ ∣ : ¬(R∧P)[x:=x̂]) ∨ Q
	 =   <Distributivity of ∨ over ∀(9.5) —¬occurs('x̂', 'Q')>
		(∀x̂ ∣ : ¬(R∧P)[x:=x̂] ∨ Q)
	 =   <Implication(3.59)>
		(∀x̂ ∣ : (R∧P)[x:=x̂] → Q)
9.31 (∃x ∣ ℤ : x+a=0)
	 Example:
	 Prove a+b = a+c → b = c
		b
	 =   <Additive identity(15.3), 0+b=b>
		0 + b
	 =   <(9.31), with witness â>
		â + a + b
	 =   <Assumption a+b=a+c>
		â + a + c
	 =   <â + a = 0 —again>
		0 + c
	 =   <Additive identity(15.3) —again>
		c
9.1 Prove that Distributivity of ∨ over ∀(9.5), P ∨ (∀x ∣ R : Q) ≡ (∀x ∣ R : P∨Q)
	(provided x does not occur free in P), follows from a similar expression with
	all ranges true : P ∨ (∀x ∣ : Q) ≡ (∀x ∣ : P∨Q) (provided x does not occur
	free in P). This means we could have used a simpler axiom.
	 Answer:
	 (P ∨ (∀x ∣ : Q) ≡ (∀x ∣ : P∨Q)) → (P ∨ (∀x ∣ R : Q) ≡ (∀x ∣ R : P∨Q))
		Assume P ∨ (∀x ∣ : Q) ≡ (∀x ∣ : P∨Q)
			(∀x ∣ R : P∨Q)
		 =   <Trading(9.3a)>
			(∀x ∣ : P∨(¬R∨Q))
		 =   <Assumption>
			P ∨ (∀x ∣ : ¬R∨Q)
		 =   <Trading(9.3a)>
			P ∨ (∀x ∣ R : Q)
9.2 Prove that (∀x ∣ R : P) ∧ (∀x ∣ R : Q) ≡ (∀x ∣ R : P∧Q) follows from a similar
	expression with all ranges true: (∀x ∣ : P) ∧ (∀x ∣ : Q) ≡ (∀x ∣ : P∧Q).
	 Answer:
	 ((∀x ∣ : P) ∧ (∀x ∣ : Q) ≡ (∀x ∣ : P∧Q)) → ((∀x ∣ R : P) ∧ (∀x ∣ R : Q) ≡ (∀x ∣ R : P∧Q))
		Assume (∀x ∣ : P) ∧ (∀x ∣ : Q) ≡ (∀x ∣ : P∧Q)
			(∀x ∣ R : P) ∧ (∀x ∣ R : Q)
		 =   <Trading(9.3a) twice>
			(∀x ∣ : ¬R∨P) ∧ (∀x ∣ : ¬R∨Q)
		 =   <Assumption>
			(∀x ∣ : (¬R∨P)∧(¬R∨Q))
		 =   <Distributivity of ∨ over ∧(3.45)>
			(∀x ∣ : ¬R∨(P∧Q))
		 =   <Trading(9.3a)>
			(∀x ∣ R : P∧Q)
9.3 Prove theorem(9.6), (∀x ∣ R : P) ≡ P ∨ (∀x ∣ : ¬R) (provided x does not occur
	free in P). Beginning with the LHS and trading seems appropriate, since the
	RHS has true as a range.
	 Answer:
		(∀x ∣ R : P)
	 =   <Trading(9.3a)>
		(∀x ∣ : ¬R∨P)
	 =   <Distributivity of ∨ over ∀(9.5)>
		P ∨ (∀x ∣ : ¬R)
9.4 Prove theorem(9.8), (∀x | R : true) ≡ true. Trading with the LHS will yield
	a formula to which some form of distributivity may be applied. Or, use (9.6).
	 Answer:
		(∀x | R : true)
	 =   <9.6>
		true ∨ (∀x ∣ : ¬R)
	 =   <Zero of ∨(3.29)>
		true
3.62 p → (q ≡ r) ≡ p ∧ q ≡ p ∧ r
9.5 Prove theorem(9.9), (∀x ∣ R : P≡Q) → ((∀x ∣ R : P) ≡ (∀x ∣ R : Q)).
	Our proof replaces the whole expression using theorem(3.62).
	 Answer:
	 (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : P) ≡ (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : Q) by Mutual Implication
	 1 (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : P) → (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : Q) by Deduction
	   1.1 (∀x ∣ R : P≡Q) Assumption
	   1.2 (∀x ∣ R : P) Assumption
	   1.3 (∀x ∣ : R→(P≡Q)) Leibniz(1.5), Trading(9.2), 1.1
	   1.4 R → (P ≡ Q) Instantiation(9.13), 1.3
	   1.5 R → P ≡ R → Q Leibniz(1.5), Distributivity of → over ≡(3.63), 1.4
	   1.6 (R → P) → (R → Q) ≡–E, 1.5
	   1.7 (∀x ∣ : R→P) Leibniz(1.5), Trading(9.2), 1.2
	   1.8 R → P Instantiation(9.13), 1.7
	   1.9 R → Q →–E, 1.8, 1.6
	   1.10 (∀x ∣ : R→Q) Metatheorem(9.16), 1.9
	   1.11 (∀x ∣ R : Q) Leibniz(1.5), Trading(9.2), 1.10
	   1.12 (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : Q) ∧–I, 1.1, 1.11
	 2 (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : Q) → ((∀x ∣ R : P≡Q) ∧ (∀x ∣ R : P)) by Deduction
	   2.1 (∀x ∣ R : P≡Q) Assumption
	   2.2 (∀x ∣ R : Q) Assumption
	   2.3 (∀x ∣ : R→(P≡Q)) Leibniz(1.5), Trading(9.2), 2.1
	   2.4 R → (P ≡ Q) Instantiation(9.13), 2.3
	   2.5 R → P ≡ R → Q Leibniz(1.5), Distributivity of → over ≡(3.63), 2.4
	   2.6 (R → P) → (R → Q) ≡–E, 2.5
	   2.7 (∀x ∣ : R→Q) Leibniz(1.5), Trading(9.2), 2.2
	   2.8 R → Q Instantiation(9.13), 2.7
	   2.9 R → P →–E, 2.8, 2.6
	   2.10 (∀x ∣ : R→P) Metatheorem(9.16), 2.9
	   2.11 (∀x ∣ R : P) Leibniz(1.5), Trading(9.2), 2.10
	   2.12 (∀x ∣ R : P≡Q) ∧ (∀x ∣ R : P) ∧–I, 2.1, 2.11
9.6 Since ∧ is idempotent, Range split for idempotent ¤(8.18) specializes for ∧
	 to (∀x ∣ R∨Q : P) ≡ (∀x ∣ R : P) ∧ (∀x ∣ Q : P)
	 However, it is possible to prove this expression without relying on axiom(8.18).
	 Develop such a proof. You may find it useful to trade R and Q into the body.
	 Answer:
		(∀x ∣ R : P) ∧ (∀x ∣ Q : P)
	 =   <Trading(9.3a), twice>
		(∀x ∣ : ¬R∨P) ∧ (∀x ∣ : ¬Q∨P)
	 =   <Distributivity(8.15)>
		(∀x ∣ : (¬R∨P)∧(¬Q∨P))
	 =   <Distributivity of ∨ over ∧(3.45)>
		(∀x ∣ : P∨(¬Q∧¬R))
	 =   <De Morgan(3.47b)>
		(∀x ∣ : ¬(Q∨R)∨P)
	 =   <Trading(9.3a)>
		(∀x ∣ R∨Q: P)
9.7 Prove Range weakening/strengthening(9.10), (∀x ∣ Q∨R : P) → (∀x ∣ Q : P).
	Range splitting may be helpful.
		(∀x ∣ Q∨R : P) → (∀x ∣ Q : P)
	 =   <Range split(8.16)>
		(∀x ∣ R : P) ∧ (∀x ∣ Q : P) → (∀x ∣ Q : P)
	 =   <Currying(3.65)>
		(∀x ∣ R : P) → ((∀x ∣ Q : P) → (∀x ∣ Q : P))
	 =   <Metatheorem(3.7) Reflexivity of →(3.71), True(3.4)>
		(∀x ∣ R : P) → true
	 =   <Right zero of ⇒(3.72))>
		true
9.8 Prove Body weakening/strengthening(9.10), (∀x ∣ R : P∧Q) → (∀x ∣ R : P).
	Distributivity of ∀ over ∧ may be helpful.
		(∀x ∣ R : P∧Q) → (∀x ∣ R : P)
	 =   <Distributivity(8.15)>
		(∀x ∣ R : Q) ∧ (∀x ∣ R : P) → (∀x ∣ R : P)
	 =   <Currying(3.65)>
		(∀x ∣ R : Q) → ((∀x ∣ R : P) → (∀x ∣ R : P))
	 =   <Metatheorem(3.7) Reflexivity of →(3.71), True(3.4)>
		(∀x ∣ R : Q) → true
	 =   <Right zero of ⇒(3.72))>
		true
9.9 Prove Monotonicity of ∀(9.12), (∀x ∣ R : Q→P) → ((∀x ∣ R : Q) → (∀x ∣ R : P)).
		(∀x ∣ R : Q→P) → ((∀x ∣ R : Q) → (∀x ∣ R : P))
	 =   <Currying(3.65)>
		(∀x ∣ R : Q→P) ∧ (∀x ∣ R : Q) → (∀x ∣ R : P)
	 (∀x ∣ R : Q→P) ∧ (∀x ∣ R : Q) → (∀x ∣ R : P) by Deduction
	 1 (∀x ∣ R : Q→P) Assumption
	 2 (∀x ∣ R : Q) Assumption
	 3 (∀x ∣ : R→(Q→P)) Leibniz(1.5), Trade(9.2), 1
	 4 (∀x ∣ : R→Q) Leibniz(1.5), Trade(9.2), 2
	 5 R → (Q → P) Instantiation(9.13), 3
	 6 R ∧ Q → P Leibniz(1.5), Currying(3.65), 5
	 7 R → P by Deduction
	   7.1 R Assumption
	   7.2 Q ∧–E, 6
	   7.3 R ∧ Q ∧–I, 8, 9
	   7.4 P →–E, 10, 6
	 8 (∀x ∣ : R→P) Metatheorem(9.16), 7
	 9 (∀x ∣ R : P) Leibniz(1.5), Trade(9.2), 8
9.10 Suppose that instead of One-point rule(8.14) for ∀, (∀x ∣ x=E : P) ≡ P[x:=E]
	 (provided x does not occur free in E), we choose the axiom
	 (∀x ∣ x=e : false) ≡ false. Prove that the one-point rule for ∀ still holds
	 (using this new axiom and theorems numbered less than(9.13))
	 Answer:
	 Assume ¬(∀x ∣ x=e : false)
		(∀x ∣ x=E : P)
	 =   <Leibniz(3.83)>
		(∀x ∣ x=E : P[x:=E])
	 =   <Identity of ∨(3.30)>
		(∀x ∣ x=E : P[x:=E]) ∨ false
	 =   <Distributivity of ∨ over ∀(9.5)>
		(∀x ∣ x=E : P[x:=E]∨false)
	 =   <Distributivity of ∨ over ∀(9.5)>
		P[x:=E] ∨ (∀x ∣ x=E : false)
	 =   <Assumption>
		P[x:=E] ∨ false
	 =   <Identity of ∨(3.30)>
		P[x:=E]
9.11 Prove Instantiation(9.13), (∀x ∣ : P) → P[x:=E]. The key is to replace
	 the dummy using Dummy renaming(8.21) so that the dummy occurs neither in
	 P nor in E.
	 Answer:
		(∀x ∣ : P) → P[x:=E]
	 =   <One-point rule(8.14)>
		(∀x ∣ : P) → (∀x ∣ x=E : P)
	 =   <Def of Implication(3.59)>
		(∀x ∣ : P) ∧ (∀x ∣ x=E : P) ≡ (∀x ∣ : P)
	 =   <Trading(9.2)>
		(∀x ∣ : P) ∧ (∀x ∣ : x=E→P) ≡ (∀x ∣ : P)
	 =   <Distributivity(8.15)>
		(∀x ∣ : P∧(x=E→P)) ≡ (∀x ∣ : P)
	 =   <(3.67)>
		(∀x ∣ : P) ≡ (∀x ∣ : P) —Reflexivity of ≡(3.5)
9.12 Prove Generalized De Morgan(9.18a), ¬(∃x ∣ R : ¬P) ≡ (∀x ∣ R : P).
	 Answer:
		¬(∃x ∣ R : ¬P)
	 =   <Generalized De Morgan(9.17)>
		¬¬(∀x ∣ R : ¬¬P)
	 =   <Double negation(3.12), twice>
		(∀x ∣ R : P)
9.13 Prove Generalized De Morgan(9.18b), ¬(∃x ∣ R : P) ≡ (∀x ∣ R : ¬P).
	 Answer:
		¬(∃x ∣ R : P)
	 =   <Generalized De Morgan(9.17)>
		¬¬(∀x ∣ R : ¬P)
	 =   <Double negation(3.12)>
		(∀x ∣ R : ¬P)
9.14 Prove Generalized De Morgan(9.18c), (∃x ∣ R : ¬P) ≡ ¬(∀x ∣ R : P).
		(∃x ∣ R : ¬P)
	 =   <Generalized De Morgan(9.17)>
		¬(∀x ∣ R : ¬¬P)
	 =   <Double negation(3.12)>
		¬(∀x ∣ R : P)
9.15 Prove Trading(9.19), (∃x ∣ R : P) ≡ (∃x ∣ : R∧P)
		(∃x ∣ R : P)
	 =   <Generalized De Morgan(9.17)>
		¬(∀x ∣ R : ¬P)
	 =   <Trading(9.3a)>
		¬(∀x ∣ : ¬R∨¬P)
	 =   <De Morgan(3.47a)>
		¬(∀x ∣ : ¬(R∧P))
	 =   <Generalized De Morgan(9.18b)>
		¬¬(∃x ∣ : R∧P)
	 =   <Double negation(3.12)>
		(∃x ∣ : R∧P)
9.16 Prove Trading(9.20), (∃x ∣ Q∧R : P) ≡ (∃x ∣ Q : R∧P)
	 Answer:
		(∃x ∣ Q∧R : P)
	 =   <Trading(9.19)>
		(∃x ∣ : (Q∧R)∧P)
	 =   <Associativity of ∧(3.37)>
		(∃x ∣ : Q∧(R∧P))
	 =   <Trading(9.19)>
		(∃x ∣ Q : R∧P)
9.17 Prove Distributivity of ∧ over ∃, (9.21), P ∧ (∃x ∣ R : Q) ≡ (∃x ∣ R : P∧Q)
	 (provided x does not occure free in P).
	 Answer:
		Assume ¬occurs('x', 'P')
			P ∧ (∃x ∣ R : Q)
		 =   <Generalized De Morgan(9.17)>
			P ∧ ¬(∀x ∣ R : ¬Q)
		 =   <Trade(9.3a)>
			P ∧ ¬(∀x ∣ R : ¬Q)
		 =   <De Morgan(3.47b)>
			¬(¬P ∨ (∀x ∣ R : ¬Q))
		 =   <Distributivity of ∨ over ∀(9.5)>
			¬(∀x ∣ R : ¬P∨¬Q)
		 =   <De Morgan(3.47a)>
			¬(∀x ∣ R : ¬(P∧Q))
		 =   <Generalized De Morgan(9.17)>
			(∃x ∣ R : P∧Q)
9.18 Prove Distributivity of ∃ over ∨, (∃x ∣ R : P) ∨ (∃x ∣ R : Q) ≡ (∃x ∣ R : P∨Q).
	 Answer:
		(∃x ∣ R : P) ∨ (∃x ∣ R : Q)
	 =   <Distributivity(8.15)>
		(∃x ∣ R : P∨Q)
9.19 Prove (9.22), (∃x ∣ R : P) ≡ P ∧ (∃x ∣ : R) (provided ¬occurs('x', 'P')).
	 Answer:
		(∃x ∣ R : P)
	 =   <Trading(9.19)>
		(∃x ∣ : R∧P)
	 =   <Distributivity of ∧ over ∃(9.21)>
		P ∧ (∃x ∣ : R)
9.20 Prove Distributivity of ∨ over ∃(9.23),
	 (∃x ∣ : R) → ((∃x ∣ R : P∨Q) ≡ P ∨ (∃x ∣ R : Q)) (provided x does not occur
	 free in P).
	 Answer:
		Assume (∃x ∣ : R)
			(∃x ∣ R : P∨Q) ≡ P ∨ (∃x ∣ R : Q)
		 =   <Trade(9.19) twice>
			(∃x ∣ : R∧(P∨Q)) ≡ P ∨ (∃x ∣ : R∧Q)
		 =   <Distributivity of ∧ over ∃(9.21), with ¬occurs('x', 'P') twice>
			(P ∨ Q) ∧ (∃x ∣ : R) ≡ P ∨ (Q ∧ (∃x ∣ : R))
		 =   <Assumption>
			(P ∨ Q) ∧ true ≡ P ∨ (Q ∧ true)
		 =   <Identity of ∧(3.39) twice>
			P ∨ Q ≡ P ∨ Q —Reflexivity of ≡(3.5)
9.21 Prove (9.24), (∃x ∣ R : false) ≡ false.
	 Answer:
		(∃x ∣ R : false)
	 =   <Trade(9.19)>
		(∃x ∣ : R∧false)
	 =   <Distributivity of ∧ over ∃(9.21)>
		false ∧ (∃x ∣ : R)
	 =   <Zero of ∧(3.40)>
		false
9.22 Since ∨ is idempotent, Range split for idempotent ¤(8.18) specializes for
	 (★x ∣ R∨S : P) = (★x ∣ R : P) ★ (★x ∣ S : P)
	 However, it is possible to prove this expression without relying on axiom(8.18).
	 Develop such a proof.
	 Answer:
		(∃x ∣ R∨S : P)
	 =   <Trading(9.19) twice>
		(∃x ∣ P : R∨S)
	 =   <Distributivity of ∨ over ∃(9.23)>
		(∃x ∣ P : R) ∨ (∃x ∣ P : S)
	 =   <Trading(9.19) four times>
		(∃x ∣ R : P) ∨ (∃x ∣ S : P)
9.23 Prove Range weakening/strengthening(9.25), (∃x ∣ R : P) → (∃x ∣ Q∨R : P).
	 Answer:
		Assume (∃x ∣ R : P)
			(∃x ∣ Q∨R : P)
		 =   <Range split(8.18)>
			(∃x ∣ Q : P) ∨ (∃x ∣ R : P)
		 =   <Assumption>
			(∃x ∣ Q : P) ∨ true
		 =   <Zero of ∨(3.29)>
			true
9.24 Prove Body weakening/strengthening(9.26), (∃x ∣ R : P) → (∃x ∣ R : P∨Q).
	 Answer:
		Assume (∃x ∣ R : P)
			(∃x ∣ R : P∨Q)
		 =   <Trading(9.19) twice>
			(∃x ∣ P∨Q : R)
		 =   <Range split(8.18)>
			(∃x ∣ P : R) ∨ (∃x ∣ Q : R)
		 =   <Trading(9.19) twice>
			(∃x ∣ R : P) ∨ (∃x ∣ Q : R)
		 =   <Assumption>
			true ∨ (∃x ∣ Q : R)
		 =   <Zero of ∨(3.29)>
			true
9.25 Prove Monotonicity of ∃(9.27), (∀x ∣ R : Q→P) → ((∃x ∣ R : Q) → (∃x ∣ R : P)).
	 Answer:
			(∀x ∣ R : Q→P) → ((∃x ∣ R : Q) → (∃x ∣ R : P))
		 =   <Metatheorem witness(9.30), –¬occurs('x̂', 'P, Q, R')>
			(∀x ∣ R : Q→P) → ((R ∧ Q)[x:=x̂] → (∃x ∣ R : P))
		 =   <Currying(3.65)>
			(∀x ∣ R : Q→P) ∧ (R ∧ Q)[x:=x̂] → (∃x ∣ R : P)
		(∀x ∣ R : Q→P) ∧ (R ∧ Q)[x:=x̂] → (∃x ∣ R : P) by Deduction
		1 (∀x ∣ R : Q→P) Assumption
		2 (R ∧ Q)[x:=x̂] Assumption
		3 (∀x ∣ : R→(Q→P)) Leibniz(1.5), Trading(9.2), 1
		4 (∀x ∣ : R∧Q→P) Leibniz(1.5), Currying(3.65), 3
		5 (R ∧ Q → P)[x:=x̂] Instantiation(9.13), 4
		6 R[x:=x̂] ∧ Q[x:=x̂] Textual Substitution, 2
		7 R[x:=x̂] ∧ Q[x:=x̂] → P[x:=x̂] Textual Substitution, 5
		8 P[x:=x̂] →–E, 6, 7
		9 R[x:=x̂] ∧–E, 6
		10 R[x:=x̂] ∧ P[x:=x̂] ∧–I, 9, 8
		11 (R ∧ P)[x:=x̂] Textual Substitution, 10
		12 (∃x ∣ R : P) ∃–I, 11
		by 9.30
		we have proved (∀x ∣ R : Q→P) ∧ (∃x ∣ R : Q) → (∃x ∣ R : P) is a theorem.
		(∀x ∣ R : Q→P) ∧ (∃x ∣ R : Q) → (∃x ∣ R : P) is a theorem
		iff 
		(∀x ∣ R : Q→P) ∧ (R ∧ Q)[x:=x̂] → (∃x ∣ R : P) is a theorem.
9.26 Prove ∃–introduction(9.28), P[x:=E] → (∃x ∣ : P).
	 Answer:
		P[x:=E] → (∃x ∣ : P)
	 =   <Contrapositive(3.61)>
		¬(∃x ∣ : P) → ¬P[x:=E]
	 =   <Generalized De Morgan(9.18b)>
		(∀x ∣ : ¬P) → ¬P[x:=E] —Instantiation(9.13)
9.27 Prove (∃x ∣ R : P) → Q ≡ (∀x ∣ R : P → Q)
	 (provided x does not occur free in Q).
	 Answer:
		(∃x ∣ R : P) → Q
	 =   <Def. of →(3.59)>
		¬(∃x ∣ R : P) ∨ Q
	 =   <Generalized De Morgan(9.18b)>
		(∀x ∣ R : ¬P) ∨ Q
	 =   <Distributivity of ∨ over ∀(9.5), with ¬occurs('x', 'Q')>
		(∀x ∣ R : ¬P∨Q)
	 =   <Def. of →(3.59)>
		(∀x ∣ R : P→Q)
9.28 Prove (∃x ∣ : R) → ((∀x ∣ R : P) → Q ≡ (∃x ∣ R : P → Q)) (provided x does
	 not occur free in P).
	 Answer:
		Assume (∃x ∣ : R)
			(∀x ∣ R : P) → Q ≡ (∃x ∣ R : P → Q)
		 =   <Generalized De Morgan(9.18a)>
			¬(∃x ∣ R : ¬P) → Q ≡ (∃x ∣ R : P → Q)
		 =   <Def. of →(3.59), Double negation(3.12)>
			(∃x ∣ R : ¬P) ∨ Q  ≡ (∃x ∣ R : P → Q)
		 =   <Trade(9.19) twice>
			(∃x ∣ : R∧¬P) ∨ Q ≡ (∃x ∣ : R∧(P→Q))
		 =   <Distributivity of ∧ over ∃(9.21), with ¬occurs('x', 'P')>
			¬P ∧ (∃x ∣ : R) ∨ Q ≡ (P → Q) ∧ (∃x ∣ : R)
		 =   <Assumption>
			¬P ∧ true ∨ Q ≡ (P → Q) ∧ true
		 =   <Identity of ∧(3.39) twice>
			¬P ∨ Q ≡ (P → Q) —Def. of →(3.59)
11.1 {x:t ∣ R : E}
	 Evaluation of the above general form of set comprehension in a state
	 yields the set of values that result from evaluating E[x:=v] in the state
	 for each value v in t such that R[x:=v] holds in that state.
	 In contexts where the type of the dummy is obvious, the type may be omitted.
	 If E has type t1, then the set comprehension has type set(t1).
11.2 {e_0, …, e_{n-1}} = {x ∣ x = e_0 ∨ … ∨ x = e_{n-1} : x}.
	 We can define a set enumeration {e_0, …, e_{n-1}} to be an abbreviation of
	 a set comprehension:
	 Examples:
	 {i ∣ 0<i<4 : i} The set {1, 2, 3}
	 {i ∣ 0<i<50∧even.i : i} Even positive integers less than 50
	 {i ∣ 0<2•i<50 : 2•i} Even positive integers less than 50
	 {x,y ∣ 1≤x≤2≤y≤3 : xʸ} The set {1², 1³, 2², 2³}
	 {x ∣ 0≤x<3 : x•y} The set {0•y, 1•y, 2•y}
	 {x ∣ 0≤x<0 : x•y} The empty set {}
Set comprehension is formalized by defining membership in the set it denotes.
For expression F:t and set {x ∣ R : E:t}(for some type t), we define:
11.3 Axiom, Set membership: Provided ¬occurs('x', 'F'),
	 F ∈ {x ∣ R : E} ≡ (∃x ∣ R : E=F).
11.4 Axiom, Extensionality: 𝘚=𝘛 ≡ (∀x ∣ : x∈𝘚≡x∈𝘛).
Table 11.1 Types of set expressions in theory set(t)
	Expression Example(with types) "Type of result"
	"Empty set, universe, variable" "∅ or 𝐔 or 𝘚" set(t)
	"Set enumeration" {e_1:t, …, e_n:t} set(t)
	"Set comprehension" {x ∣ R:𝔹 : E:t}{x:t ∣ R:𝔹} set(t)
	"Set membership" x:t∈𝘚:set(t) 𝔹
	"Set equality" 𝘚:set(t) = 𝘚:set(t) 𝔹
	"Set size' # 𝘚:set(t) ℕ
	⊂, ⊃, ⊆, ⊇ 𝘚:set(t)⊆𝘚:set(t) 𝔹
	Complement ~𝘚:set(t) set(t)
	⋃, ⋂, ¬ 𝘚:set(t)⋃𝘚:set(t) set(t)
	"Power set" (𝒫𝘚):set(t) set(set(t))
11.5 𝘚 = {x ∣ x∈𝘚 : x}
	 According to axiom Extensionality(11.4), it suffices to prove that
	 v∈𝘚 ≡ v∈{x ∣ x∈𝘚: x}, for arbitrary v. We have,
		v ∈ {x ∣ x∈𝘚: x}
	 =   <Def. of membership(11.3)>
		(∃x ∣ x∈𝘚 : v=x)
	 =   <Trading(9.19) twice>
		(∃x ∣ x=v : x∈𝘚)
	 =   <One-point rule(8.14)>
		v∈𝘚
Traditional form of set comprehension
{x ∣ R}
(x is a single variable), which we view as an abbreviation of {x ∣ R : x}.
Example:
{x+y ∣ x=y+1}
	If x is the dummy, then the set is {2•y+1};
		{y+1+y}
	If y is the dummy, then the set is
		{x-1+x}
	If x and y are the dummies, then it is the set of odd integers.
		{x+y=y+1+y}
			{y+1+y ∣ y:ℕ}
11.6 Provided ¬occurs('y', 'R') and ¬occurs('y', 'E'),
	 {x ∣ R : E} = {y ∣ (∃x ∣ R : E=y)}.
Following theorem formalizes the connection between sets and predicates:
a predicate is a representation for the set of argument-values for which it is true.
11.7 x ∈ {x ∣ R} ≡ R
	 Note thta x is used with two different meanings in the LHS of (11.7).
	 The leftmost occurrence of x is free, as are free occurrences of x in the
	 RHS. All occurrences of x in {x ∣ R} are bound. Since (11.7) is valid, by
	 instantiating free variable x with any expression y we have
	 y∈{x ∣ R} ≡ R[x:=y] for any expression y.
11.8 Principle of comprehension. To each predicate R, there corresponds a set
	 comprehension {x:t ∣ R}, which contains the objects in t that satisfy R;
	 R is called a characteristic predicate of the set.
	𝘚 = {3, 5}
	𝘚 = {x ∣ x=3∨x=5}
	x∈𝘚 ≡ x=3 ∨ x=5 (for all x).
11.9 {x ∣ Q} = {x ∣ R} ≡ (∀x ∣ : Q≡R)
11.10 Theorem. {x ∣ Q} = {x ∣ R} is valid iff Q≡R is valid.
11.11 Methods for proving set equality 𝘚=𝘛:
	  (a) Use Leibniz directly.
	  (b) Use axiom Extensionality(11.4) and prove v∈𝘚 ≡ v∈𝘛
		  for an arbitrary value v.
	  (c) Prove Q ≡ R and conclude {x ∣ Q}={x ∣ R}.
11.12 Axiom, Size: #𝘚 = (∑x ∣ x∈𝘚 : 1)
	  Could be written as
		(∀S:set(t) ∣ : #𝘚=(∑x ∣ x∈𝘚 : 1))
		Metatheorem(9.16)
11.13 Axiom, Subset: 𝘚⊆𝘛 ≡ (∀x ∣ x∈𝘚 : x∈𝘛)
11.14 Axiom, Proper subset: 𝘚⊂𝘛 ≡ 𝘚⊆𝘛 ∧ 𝘚≠𝘛
11.15 Axiom, Superset: 𝘚⊇𝘛 ≡ 𝘛⊆𝘚
11.16 Axiom, Proper superset: 𝘚⊃𝘛 ≡ 𝘛⊂𝘚
11.17 Axiom, Complement: v∈~𝘚 ≡ v∈𝐔 ∧ v∉𝘚
	  Example:
		𝐔 = {0, 1, 2, 3, 4, 5}, we have
			~{3, 5} = {0, 1, 2, 4},
			~𝐔 = ∅, ~∅ = 𝐔.
11.18 v∈~𝘚 ≡ v∉𝘚 (for v in 𝐔)
11.19 ~~𝘚 = 𝘚
11.20 Axiom, Union: v∈𝘚∪𝘛 ≡ v∈𝘚 v v∈𝘛
11.21 Axiom, Intersection: v∈𝘚∩𝘛 ≡ v∈𝘚 ∧ v∈𝘛
11.22 Axiom, Difference: v∈𝘚-𝘛 ≡ v∈𝘚 ∧ v∉𝘛
	  Example:
		{3, 5, 6}∪{3, 2, 1}={3, 5, 6, 2, 1},
		{3, 5, 6}∩{3, 2, 1}={3},
		{3, 5, 6}-{3, 2, 1}={5, 6}.
11.23 Axiom, Power set: v∈𝒫𝘚 ≡ v⊆𝘚
	  Example:
		𝒫{3, 5} ≡ {∅, {3}, {5}, {3,5}}
11.24 Definition. Let Eₛ be a set expression constructed from set variables, ∅,
	  𝐔(a universe for all set variables in question), ~, ∪ and ∩.
	  Then Eₚ is the expression constructed from Eₛ by replacing
		∅ with false, 𝐔 with true,
		∪ with ∨, ∩ with ∧,
		~ with ¬.
		The construction is reversible: Eₛ can be constructed from Eₚ.
11.25 Metatheorem. For any set expressions Eₛ and Fₛ:
		(a) Eₛ=Fₛ is valid iff Eₚ ≡ Fₚ is valid,
		(b) Eₛ⊆Fₛ is valid iff Eₚ → Fₚ is valid,
		(c) Eₛ=𝐔 is valid iff Eₚ is valid.
11.26 Symmetry of ∪: 𝘚∪𝘛=𝘛∪𝘚
11.27 Associativity of ∪: (𝘚∪𝘛)∪𝑈=𝘚∪(𝘛∪𝑈)
11.28 Idempotency of ∪: 𝘚∪𝘚=𝘚
11.29 Zero of ∪: 𝘚∪𝐔=𝐔
11.30 Identity of ∪: 𝘚∪∅=𝘚
11.31 Weakening: 𝘚⊆𝘚∪𝘛
11.32 Excluded middle: 𝘚∪~𝘚=𝐔
11.33 Symmetry of ∩: S∩𝘛=𝘛∩S
11.34 Associativity of ∩: (𝘚∩𝘛)∩𝑈=𝘚∩(𝘛∩𝑈)
11.35 Idempotency of ∩: 𝘚∩𝘚=𝘚
11.36 Zero of ∩: 𝘚∩∅=∅
11.37 Identity of ∩: 𝘚∩𝐔=𝘚
11.38 Strengthening: 𝘚∩𝘛⊆𝘚
11.39 Contradiction: 𝘚∩~𝘚=∅
11.40 Distributivity of ∪ over ∩:
	  𝘚∪(𝘛∩𝑈)=(𝘚∪𝘛)∩(𝘚∪𝑈)
11.41 Distributivity of ∩ over ∪:
	  𝘚∩(𝘛∪𝑈)=(𝘚∩𝘛)∪(𝘚∩𝑈)
11.42 De Morgan: (a) ~(𝘚∪𝘛)=~𝘚∩~𝘛
				 (b) ~(𝘚∩𝘛)=~𝘚∪~𝘛
11.43 𝘚⊆𝘛 ∧ 𝑈⊆𝑉 → (𝘚∪𝑈)⊆(𝘛∪𝑉)
11.44 𝘚⊆𝘛 ∧ 𝑈⊆𝑉 → (𝘚∩𝑈)⊆(𝘛∩𝑉)
11.45 𝘚⊆𝘛 ≡ 𝘚∪𝘛=𝘛
11.46 𝘚⊆𝘛 ≡ 𝘚∩𝘛=𝘚
11.47 𝘚∪𝘛=𝐔 ≡ (∀x ∣ x∈𝐔 : x∉𝘚→x∈𝘛)
11.48 𝘚∩𝘛=∅ ≡ (∀x ∣ : x∈𝘚→x∉𝘛)
11.49 𝘚-𝘛=𝘚∩~𝘛
11.50 𝘚-𝘛⊆𝘚
11.51 𝘚-∅=𝘚
11.52 𝘚∩(𝘛-𝘚)=∅
11.53 𝘚∪(𝘛-𝘚)=𝘚∪𝘛
11.54 𝘚-(𝘛∪𝑈)=(𝘚-𝘛)∩(𝘚-𝑈)
11.55 𝘚-(𝘛∩𝑈)=(𝘚-𝘛)∪(𝘚-𝑈)
11.56 (∀x ∣ : P→Q) ≡ {x ∣ P}⊆{x ∣ Q}
	  Proof:
		{x ∣ P} ⊆ {x ∣ Q}
	 =   <Subset ⊆(11.13), where ¬occurs('v', 'P, Q')>
		(∀v ∣ v∈{x ∣ P} : v∈{x ∣ Q})
	 =   <(11.7) twice>
		(∀v ∣ P[x:=v] : Q[x:=v])
	 =   <Trading(9.2); Dummy renaming(8.21)>
		(∀x ∣ : P[x:=v][v:=x]→Q[x:=v][v:=x])
	 =   <Property of textual substitution, ¬occurs('v', 'P, Q')>
		(∀x ∣ : P→Q)
11.57 Antisymmetry: 𝘚⊆𝘛 ∧ 𝘛⊆𝘚 ≡ 𝘚=𝘛
11.58 Reflexivity: 𝘚⊆𝘚
11.59 Transitivity: 𝘚⊆𝘛 ∧ 𝘛⊆𝑈 → 𝘚⊆𝑈
11.60 ∅⊆𝘚
11.61 𝘚⊂𝘛 ≡ 𝘚⊆𝘛 ∧ ¬(𝘛⊆𝘚)
11.62 𝘚⊂𝘛 ≡ 𝘚⊆𝘛 ∧ (∃x ∣ x∈𝘛 : x∉𝘚)
	  ¬(𝘛⊆𝘚)
	  (∀x ∣ x∈𝘛 : x∈𝘚)
	  (∃x ∣ x∈𝘛 : ¬(x∈𝘚))
11.63 𝘚⊆𝘛 ≡ 𝘚⊂𝘛 ∨ 𝘚=𝘛
11.64 𝘚⊄𝘚
11.65 𝘚⊂𝘛 → 𝘚⊆𝘛
	  S is a proper subset of T implies S is a subset of T.
	  S is a subset of T does not implies S is a proper subset of T.
11.66 𝘚⊂𝘛 → 𝘛⊈𝘚
11.67 𝘚⊆𝘛 → 𝘛⊄𝘚
	  S is a proper subset of T implies T is not a subset of S.
	  S is a subset of T implies T is not a proper subset of S.
11.68 𝘚⊆𝘛 ∧ ¬(𝑈⊆𝘛) → ¬(𝑈⊆𝘚)
	  S is a subset of T and U is not a subset of T
	  implies U is not a subset of S.
	  Proof.
		𝘚⊆𝘛 ∧ ¬(𝑈⊆𝘛)
	 =   <Subset ⊆(11.13); Generalized De Morgan(9.18c)>
		𝘚⊆𝘛 ∧ (∃x ∣ x∈𝑈 : x∉𝘛)
	 =   <Distributivity of ∧ over ∃(9.21)>
		(∃x ∣ x∈𝑈 : x∉𝘛∧𝘚⊆𝘛)
	 =   <Subset ⊆(11.13); Trading(9.2)>
		(∃x ∣ x∈𝑈 : x∉𝘛∧(∀x ∣ : x∈𝘚→x∈𝘛))
	 =   <Monotonicity of ∃(9.27), using Instantiation(9.13)>
		()
	 =   <>
		
11.69 (∃x ∣ x∈𝘚 : x∉𝘛) → 𝘚≠𝘛
11.70 Transitivity: (a) 𝘚⊆𝘛 ∧ 𝘛⊂𝑈 → 𝘚⊂𝑈
					(b) 𝘚⊂𝘛 ∧ 𝘛⊆𝑈 → 𝘚⊂𝑈
					(c) 𝘚⊂𝘛 ∧ 𝘛⊂𝑈 → 𝘚⊂𝑈
3.82 Transitivity:
	(a) (p → q) ∧ (q → r) → (p → r)
	(b) (p ≡ q) ∧ (q → r) → (p → r)
	(c) (p → q) ∧ (q ≡ r) → (p → r)