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forallx-do-sol-prooffol.tex
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%!TEX root = forallxsol.tex
%\part{Natural deduction for FOL}
%\label{ch.NDFOL}
%\addtocontents{toc}{\protect\mbox{}\protect\hrulefill\par}
\chapter{Grundregeln für die LEO}\setcounter{ProbPart}{0}
\problempart
Erklären Sie, wieso die folgenden Beweise \emph{inkorrekt} sind. Geben Sie auch Interpretationen an, die zeigen, dass die den Beweisen zugrundeliegenden Argumente ungültig sind:
\begin{multicols}{2}
\begin{fitchproof}
\hypo{Rxx}{\forall x\,\atom{R}{x,x}}
\have{Raa}{\atom{R}{a,a}}\Ae{Rxx}
\have{Ray}{\forall y\,\atom{R}{a,y}}\Ai{Raa}
\have{Rxy}{\forall x \forall y\,\atom{R}{x,y}}\Ai{Ray}
\end{fitchproof}
\vfill
\noindent\myanswer{Wenn wir $\forall$I nutzen, müssen wir \emph{alle Vorkommnisse} eines Namens mit der neuen Variable ersetzen. Also ist Zeile 3 unzulässig.}
\myanswer{Hier ist eine Gegeninterpretation:
\begin{center}
\begin{tikzpicture}
\node (atom4) at (0,0) {1};
\node (atom5) at (2,0) {2};
\draw[->, thick] (atom4)+(-0.15,0.15) arc (-330:-30:.3);
\draw[->, thick] (atom5)+(0.15,-0.15) arc (-150:150:.3);
\end{tikzpicture}
\end{center}}
\columnbreak
\begin{fitchproof}
\hypo{AE}{\forall x \exists y\,\atom{R}{x,y}}
\have{E}{\exists y\,\atom{R}{a,y}}\Ae{AE}
\open
\hypo{ass}{\atom{R}{a,a}}
\have{Ex}{\exists x\,\atom{R}{x,x}}\Ei{ass}
\close
\have{con}{\exists x\,\atom{R}{x,x}}\Ee{E, ass-Ex}
\end{fitchproof}
\vfill
\noindent\myanswer{Der ersetzende Name `$a$', kommt bereits in Zeile 2 vor, auf die wir versuchen $\exists$E anzuwenden. Also ist die Nutzung von $\exists$E in Zeile 5 unzulässig.}
\myanswer{Hier ist eine Gegeninterpretation:
\begin{center}
\begin{tikzpicture}
\node (atom4) at (0,0) {1};
\node (atom5) at (2,0) {2};
\draw[<->, thick] (atom4)--(atom5);
\end{tikzpicture}
\end{center}}
\end{multicols}
\newpage
\problempart
\label{pr.justifyFOLproof}
Den folgenden drei Beweisen fehlen Zitationen (Regeln und Zeilennummern). Fügen Sie diese hinzu.
\begin{earg}
\item\begin{fitchproof}
\hypo{p1}{\forall x\exists y(\atom{R}{x,y} \eor \atom{R}{y,x})}
\hypo{p2}{\forall x\enot \atom{R}{m,x}}
\have{3}{\exists y(\atom{R}{m,y} \eor \atom{R}{y,m})}\Ae{p1}
\open
\hypo{a1}{\atom{R}{m,a} \eor \atom{R}{a,m}}
\have{a2}{\enot \atom{R}{m,a}}\Ae{p2}
\have{a3}{\atom{R}{a,m}}\ds{a1, a2}
\have{a4}{\exists x\,\atom{R}{x,m}}\Ei{a3}
\close
\have{n}{\exists x\,\atom{R}{x,m}}\Ee{3, a1-a4}
\end{fitchproof}
\item\begin{fitchproof}
\hypo{1}{\forall x(\exists y\,\atom{L}{x,y} \eif \forall z\atom{L}{z,x})}
\hypo{2}{\atom{L}{a,b}}
\have{3}{\exists y\,\atom{L}{a,y} \eif \forall z\atom{L}{z,a}}\Ae{1}
\have{4}{\exists y\,\atom{L}{a,y}}\Ei{2}
\have{5}{\forall z \atom{L}{z,a}}\ce{3, 4}
\have{6}{\atom{L}{c,a}}\Ae{5}
\have{7}{\exists y\,\atom{L}{c,y} \eif \forall z\atom{L}{z,c}}\Ae{1}
\have{8}{\exists y\,\atom{L}{c,y}}\Ei{6}
\have{9}{\forall z \atom{L}{z,c}}\ce{7, 8}
\have{10}{\atom{L}{c,c}}\Ae{9}
\have{11}{\forall x\,\atom{L}{x,x}}\Ai{10}
\end{fitchproof}
\newpage
\item\begin{fitchproof}
\hypo{a}{\forall x(\atom{J}{x} \eif \atom{K}{x})}
\hypo{b}{\exists x\forall y\,\atom{L}{x,y}}
\hypo{c}{\forall x\,\atom{J}{x}}
\open
\hypo{2}{\forall y\,\atom{L}{a,y}}
\have{3}{\atom{L}{a,a}}\Ae{2}
\have{d}{\atom{J}{a}}\Ae{c}
\have{e}{\atom{J}{a} \eif \atom{K}{a}}\Ae{a}
\have{f}{\atom{K}{a}}\ce{e, d}
\have{4}{\atom{K}{a} \eand \atom{L}{a,a}}\ai{f, 3}
\have{5}{\exists x(\atom{K}{x} \eand \atom{L}{x,x})}\Ei{4}
\close
\have{j}{\exists x(\atom{K}{x} \eand \atom{L}{x,x})}\Ee{b, 2-5}
\end{fitchproof}
\end{earg}
\problempart
\label{pr.BarbaraEtc.proof1}
In \S\ref{s:MoreMonadic} Übung A betrachteten wir 15 syllogistische Figuren. Geben Sie Beweise, die diesen Argumentformen entsprechen. Beachten Sie: Die Übung ist \emph{viel} leichter, wenn Sie `Kein F ist G' (beispielsweise) als `$\forall x (\atom{F}{x} \eif \enot \atom{G}{x})$' symbolisieren.
\myanswer{Wir beweisen die ersten vier syllogistischen Figuren; die verbleibenden Beweise sind extrem ähnlich.}
\myanswer{\noindent \textbf{Barbara}
\begin{fitchproof}
\hypo{gf}{\forall x (\atom{G}{x} \eif \atom{F}{x})}
\hypo{hg}{\forall x (\atom{H}{x} \eif \atom{G}{x})}
\have{gafa}{\atom{G}{a} \eif \atom{F}{a}}\Ae{gf}
\have{haga}{\atom{H}{a} \eif \atom{G}{a}}\Ae{hg}
\open
\hypo{ha}{\atom{H}{a}}
\have{ga}{\atom{G}{a}}\ce{haga, ha}
\have{fa}{\atom{F}{a}}\ce{gafa, ga}
\close
\have{hafa}{\atom{H}{a} \eif \atom{F}{a}}\ci{ha-fa}
\have{con}{\forall x (\atom{H}{x} \eif \atom{F}{x})}\Ai{hafa}
\end{fitchproof}}
\myanswer{\noindent\textbf{Celerant} funktioniert genauso wie Barbara, wenn wir `$F$' durchgehend mit `$\enot F$' ersetzen.}
\newpage
\myanswer{\textbf{Ferio}
\begin{fitchproof}
\hypo{gnf}{\forall x (\atom{G}{x} \eif \enot \atom{F}{x})}
\hypo{hg}{\exists x (\atom{H}{x} \eand \atom{G}{x})}
\open
\hypo{haga}{\atom{H}{a} \eand \atom{G}{a}}
\have{ha}{\atom{H}{a}}\ae{haga}
\have{ga}{\atom{G}{a}}\ae{haga}
\have{ganfa}{\atom{G}{a} \eif \enot \atom{F}{a}}\Ae{gnf}
\have{nfa}{\enot \atom{F}{a}}\ce{ganfa, ga}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ee{hg, haga-hnf}
\end{fitchproof}}
\myanswer{\textbf{Darii} funktioniert genauso wie Ferio, wenn wir `$\enot F$' durchgehend mit `$F$' ersetzen.}
\
\problempart
\label{pr.BarbaraEtc.proof2}
Aristoteles und seine Nachfolger:innen identifizierten weitere syllogistische Formen. Symbolisieren Sie diese in der LEO und geben Sie dementsprechende Beweise.
\begin{earg}
\item \textbf{Barbari.} Etwas ist H. Alle Gs sind F. Alle Hs sind G. Also: Ein H ist F
\myanswer{$$\exists x\,\atom{H}{x}, \forall x(\atom{G}{x} \eif \atom{F}{x}), \forall x (\atom{H}{x} \eif \atom{G}{x}) \therefore \exists x (\atom{H}{x} \eand \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{H}{x}}
\hypo{gf}{\forall x (\atom{G}{x} \eif \atom{F}{x})}
\hypo{hg}{\forall x (\atom{H}{x} \eif \atom{G}{x})}
\open
\hypo{ha}{\atom{H}{a}}
\have{haga}{\atom{H}{a} \eif \atom{G}{a}}\Ae{hg}
\have{ga}{\atom{G}{a}}\ce{haga, ha}
\have{gafa}{\atom{G}{a} \eif \atom{F}{a}}\Ae{gf}
\have{fa}{\atom{F}{a}}\ce{gafa, ga}
\have{hafa}{\atom{H}{a} \eand \atom{F}{a}}\ai{ha, fa}
\have{hf}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ei{hafa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ee{imp, ha-hf}
\end{fitchproof}}
\item \textbf{Celaront.} Etwas ist H. Kein G ist F. Alle Hs sind G. Also: Ein H ist nicht F
\myanswer{$$\exists x\,\atom{H}{x}, \forall x(\atom{G}{x} \eif \enot \atom{F}{x}), \forall x (\atom{H}{x} \eif \atom{G}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
Beweis funktioniert genauso wie der für Barbari, wenn wir `$F$' durchgehend mit `$\enot F$' ersetzen.}
\item \textbf{Cesaro.} Etwas ist H. Kein F ist G. Alle Hs sind G. Also: Ein H ist nicht F.
\myanswer{$$\exists x\,\atom{H}{x}, \forall x(\atom{F}{x} \eif \enot \atom{G}{x}), \forall x (\atom{H}{x} \eif \atom{G}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{H}{x}}
\hypo{fg}{\forall x (\atom{F}{x} \eif \enot \atom{G}{x})}
\hypo{hng}{\forall x (\atom{H}{x} \eif \atom{G}{x})}
\open
\hypo{ha}{\atom{H}{a}}
\have{hanga}{\atom{H}{a} \eif \atom{G}{a}}\Ae{hng}
\have{nga}{\atom{G}{a}}\ce{hanga, ha}
\have{faga}{\atom{F}{a} \eif \enot \atom{G}{a}}\Ae{fg}
\open
\hypo{fa}{\atom{F}{a}}
\have{ga}{\enot \atom{G}{a}}\ce{faga, fa}
\have{red}{\ered}\ri{nga, ga}
\close
\have{nfa}{\enot \atom{F}{a}}\ni{fa-red}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ee{imp, ha-hnf}
\end{fitchproof}}
\item \textbf{Camestros.} Etwas ist H. Alle Fs sind G. Kein H ist G. Also: Ein H ist nicht F.
\myanswer{$$\exists x\,\atom{H}{x}, \forall x(\atom{F}{x} \eif \atom{G}{x}), \forall x (\atom{H}{x} \eif \enot \atom{G}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{H}{x}}
\hypo{fg}{\forall x (\atom{F}{x} \eif \atom{G}{x})}
\hypo{hng}{\forall x (\atom{H}{x} \eif \enot \atom{G}{x})}
\open
\hypo{ha}{\atom{H}{a}}
\have{hanga}{\atom{H}{a} \eif \enot \atom{G}{a}}\Ae{hng}
\have{nga}{\enot \atom{G}{a}}\ce{hanga, ha}
\have{faga}{\atom{F}{a} \eif \atom{G}{a}}\Ae{fg}
\have{nfa}{\enot \atom{F}{a}}\mt{faga, nga}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ee{imp, ha-hnf}
\end{fitchproof}}
\item \textbf{Felapton.} Etwas ist G. Kein G ist F. Alle Gs sind H. Also: Ein H ist nicht F.
\myanswer{$$\exists x\,\atom{G}{x}, \forall x (\atom{G}{x} \eif \enot \atom{F}{x}), \forall x(\atom{G}{x} \eif \atom{H}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{G}{x}}
\hypo{gnf}{\forall x (\atom{G}{x} \eif \enot \atom{F}{x})}
\hypo{gh}{\forall x (\atom{G}{x} \eif \atom{H}{x})}
\open
\hypo{ga}{\atom{G}{a}}
\have{gaha}{\atom{G}{a} \eif \atom{H}{a}}\Ae{gh}
\have{ha}{\atom{H}{a}}\ce{gaha, ga}
\have{ganfa}{\atom{G}{a} \eif \enot \atom{F}{a}}\Ae{gnf}
\have{nfa}{\enot \atom{F}{a}}\ce{ganfa, ga}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ee{imp, ga-hnf}
\end{fitchproof}}
\item \textbf{Darapti.} Etwas ist G. Alle Gs sind F. Alle Gs sind H. Also: Ein H ist F.
\myanswer{$$\exists x\,\atom{G}{x}, \forall x (\atom{G}{x} \eif \atom{F}{x}), \forall x(\atom{G}{x} \eif \atom{H}{x}) \therefore \exists x (\atom{H}{x} \eand \atom{F}{x})$$
Beweis funktioniert genauso wie für Felapton, wenn wir `$\enot F$' durchgehend mit `$F$' ersetzen.}
\newpage \item \textbf{Calemos.} Etwas ist H. Alle Fs sind G. Kein G ist H. Also: Ein H ist nicht F.
\myanswer{$$\exists x\,\atom{H}{x}, \forall x(\atom{F}{x} \eif \atom{G}{x}), \forall x(\atom{G}{x} \eif \enot \atom{H}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{H}{x}}
\hypo{fg}{\forall x (\atom{F}{x} \eif \atom{G}{x})}
\hypo{gnh}{\forall x (\atom{G}{x} \eif \enot \atom{H}{x})}
\open
\hypo{ha}{\atom{H}{a}}
\have{ganha}{\atom{G}{a} \eif \enot \atom{H}{a}}\Ae{gnh}
\open
\hypo{ga}{\atom{G}{a}}
\have{nha}{\enot \atom{H}{a}}\ce{ganha, ga}
\have{red}{\ered}\ri{ha, nha}
\close
\have{nga}{\enot \atom{G}{a}}\ni{ga-red}
\have{faga}{\atom{F}{a} \eif \atom{G}{a}}\Ae{fg}
\have{nfa}{\enot \atom{F}{a}}\mt{faga, nga}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ee{imp, ha-hnf}
\end{fitchproof}}
\newpage
\item \textbf{Fesapo.} Etwas ist G. Kein F ist G. All Gs sind H. Also: Ein H ist nicht F.
\myanswer{$$\exists x\,\atom{G}{x}, \forall x(\atom{F}{x} \eif \enot \atom{G}{x}), \forall x(\atom{G}{x} \eif \atom{H}{x}) \therefore \exists x (\atom{H}{x} \eand \enot \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{G}{x}}
\hypo{fng}{\forall x (\atom{F}{x} \eif \enot \atom{G}{x})}
\hypo{gh}{\forall x (\atom{G}{x} \eif \atom{H}{x})}
\open
\hypo{ga}{\atom{G}{a}}
\have{gaha}{\atom{G}{a} \eif \atom{H}{a}}\Ae{gh}
\have{ha}{\atom{H}{a}}\ce{gaha, ga}
\have{fanga}{\atom{F}{a} \eif \enot \atom{G}{a}}\Ae{fng}
\open
\hypo{fa}{\atom{F}{a}}
\have{nga}{\enot \atom{G}{a}}\ce{fanga, fa}
\have{red}{\ered}\ri{ga, nga}
\close
\have{nfa}{\enot \atom{F}{a}}\ni{fa-red}
\have{hanfa}{\atom{H}{a} \eand \enot \atom{F}{a}}\ai{ha, nfa}
\have{hnf}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ei{hanfa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \enot \atom{F}{x})}\Ee{imp, ga-hnf}
\end{fitchproof}}
\newpage
\item \textbf{Bamalip.} Etwas ist F. Alle Fs sind G. Alle Gs sind H. Also: Ein H ist F.
\myanswer{$$\exists x\,\atom{F}{x}, \forall x(\atom{F}{x} \eif \atom{G}{x}), \forall x(\atom{G}{x} \eif \atom{H}{x}) \therefore \exists x (\atom{H}{x} \eand \atom{F}{x})$$
\begin{fitchproof}
\hypo{imp}{\exists x\,\atom{F}{x}}
\hypo{fg}{\forall x (\atom{F}{x} \eif \atom{G}{x})}
\hypo{gh}{\forall x (\atom{G}{x} \eif \atom{H}{x})}
\open
\hypo{fa}{\atom{F}{a}}
\have{faga}{\atom{F}{a} \eif \atom{G}{a}}\Ae{fg}
\have{ga}{\atom{G}{a}}\ce{faga, fa}
\have{gaha}{\atom{G}{a} \eif \atom{H}{a}}\Ae{gh}
\have{ha}{\atom{H}{a}}\ce{gaha, ga}
\have{hafa}{\atom{H}{a} \eand \atom{F}{a}}\ai{ha, fa}
\have{hf}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ei{hafa}
\close
\have{con}{\exists x (\atom{H}{x} \eand \atom{F}{x})}\Ee{imp, fa-hf}
\end{fitchproof}}
\end{earg}
\problempart
\label{pr.someFOLproofs}
Beweisen Sie jede der folgenden Aussagen.
\begin{earg}
\item $\proves \forall x\,\atom{F}{x} \eif \forall y(\atom{F}{y} \eand \atom{F}{y})$
\myanswer{\begin{fitchproof}
\open
\hypo{Af}{\forall x\,\atom{F}{x}}
\have{con}{\atom{F}{a}}\Ae{Af}
\have{con2}{\atom{F}{a} \eand \atom{F}{a}}\ai{con,con}
\have{con3}{\forall y\,(\atom{F}{y} \eand \atom{F}{y})}\Ai{con2}
\close
\have{con4}{\forall x\,\atom{F}{x} \eif \forall y(\atom{F}{x} \eand \atom{F}{x})}\ci{Af-con3}
\end{fitchproof}}
\newpage
\item $\forall x(Ax\eif \atom{B}{x}), \exists x\,\atom{A}{x} \proves \exists x\,\atom{B}{x}$
\myanswer{\begin{fitchproof}
\hypo{Aab}{\forall x (\atom{A}{x} \eif \atom{B}{x})}
\hypo{Ea}{\exists x\,\atom{A}{x}}
\open
\hypo{a}{\atom{A}{a}}
\have{ab}{\atom{A}{a} \eif \atom{B}{a}}\Ae{Aab}
\have{b}{\atom{B}{a}}\ce{ab, a}
\have{Eb}{\exists x\,\atom{B}{x}}\Ei{b}
\close
\have{Eb1}{\exists x\,\atom{B}{x}}\Ee{Ea, a-Eb}
\end{fitchproof}}
\item $\forall x(\atom{M}{x} \eiff \atom{N}{x}), \atom{M}{a} \eand \exists x\,\atom{R}{x,a}\proves \exists x\,\atom{N}{x}$
\myanswer{\begin{fitchproof}
\hypo{Amn}{\forall x (\atom{M}{x} \eiff \atom{N}{x})}
\hypo{mEr}{\atom{M}{a} \eand \exists x\,\atom{R}{x,a}}
\have{m}{\atom{M}{a}}\ae{mEr}
\have{mn}{\atom{M}{a} \eiff \atom{N}{a}}\Ae{Amn}
\have{n}{\atom{N}{a}}\be{mn, m}
\have{En}{\exists x\,\atom{N}{x}}\Ei{n}
\end{fitchproof}}
\item $\forall x \forall y\,\atom{G}{x,y}\proves\exists x\,\atom{G}{x,x}$
\myanswer{\begin{fitchproof}
\hypo{AAg}{\forall x \forall y\,\atom{G}{x,y}}
\have{Ag}{\forall y\,\atom{G}{a,y}}\Ae{AAg}
\have{g}{\atom{G}{a,a}}\Ae{Ag}
\have{Eg}{\exists x\,\atom{G}{x,x}}\Ei{g}
\end{fitchproof}}
\newpage
\item $\proves\forall x\,\atom{R}{x,x}\eif \exists x \exists y\,\atom{R}{x,y}$
\myanswer{\begin{fitchproof}
\open
\hypo{Ar}{\forall x\,\atom{R}{x,x}}
\have{r}{\atom{R}{a,a}}\Ae{Ar}
\have{Er}{\exists y\,\atom{R}{a,y}}\Ei{r}
\have{EEr}{\exists x \exists y\,\atom{R}{x,y}}\Ei{Er}
\close
\have{ArEEr}{\forall x\,\atom{R}{x,x} \eif \exists x \exists y\,\atom{R}{x,y}}\ci{Ar-EEr}
\end{fitchproof}}
\item $\proves\forall y \exists x (\atom{Q}{y} \eif \atom{Q}{x})$
\myanswer{\begin{fitchproof}
\open
\hypo{q}{\atom{Q}{a}}
\have{q1}{\atom{Q}{a}}\by{R}{q}
\close
\have{qq}{\atom{Q}{a} \eif \atom{Q}{a}}\ci{q-q1}
\have{Eqq}{\exists x(\atom{Q}{a} \eif \atom{Q}{x})}\Ei{qq}
\have{AEqq}{\forall y \exists x(\atom{Q}{y} \eif \atom{Q}{x})}\Ai{Eqq}
\end{fitchproof}}
\item $Na \eif \forall x(\atom{M}{x} \eiff \atom{M}{a}), \atom{M}{a}, \enot \atom{M}{b}\proves \enot \atom{N}{a}$
\myanswer{\begin{fitchproof}
\hypo{nAmm}{\atom{N}{a} \eif \forall x(\atom{M}{x} \eiff \atom{M}{a})}
\hypo{m}{\atom{M}{a}}
\hypo{nm}{\enot \atom{M}{b}}
\open
\hypo{n}{\atom{N}{a}}
\have{Amm}{\forall x (\atom{M}{x} \eiff \atom{M}{a})}\ce{nAmm, n}
\have{mm}{\atom{M}{b} \eiff \atom{M}{a}}\Ae{Amm}
\have{mb}{\atom{M}{b}}\be{mm, m}
\have{red}{\ered}\ri{mb, nm}
\close
\have{nn}{\enot \atom{N}{a}}\ni{n-red}
\end{fitchproof}}
\newpage \item $\forall x \forall y (\atom{G}{x,y} \eif \atom{G}{y,x}) \proves \forall x\forall y (\atom{G}{x,y} \eiff \atom{G}{y,x})$
\myanswer{\begin{fitchproof}
\hypo{AAgg}{\forall x \forall y(\atom{G}{x,y} \eif \atom{G}{y,x})}
\open
\hypo{gab}{\atom{G}{a,b}}
\have{Agg}{\forall y(\atom{G}{a,y} \eif \atom{G}{y,a})}\Ae{AAgg}
\have{gg}{\atom{G}{a,b} \eif \atom{G}{b,a}}\Ae{Agg}
\have{gba}{\atom{G}{b,a}}\ce{gg, gab}
\close
\open
\hypo{gba1}{\atom{G}{b,a}}
\have{Agg1}{\forall y(\atom{G}{b,y} \eif \atom{G}{y,b})}\Ae{AAgg}
\have{gg1}{\atom{G}{b,a} \eif \atom{G}{a,b}}\Ae{Agg1}
\have{gab1}{\atom{G}{a,b}}\ce{gg1, gba1}
\close
\have{bi}{\atom{G}{a,b} \eiff \atom{G}{b,a}}\bi{gab-gba, gba1-gab1}
\have{Abi}{\forall y(\atom{G}{a,y} \eiff \atom{G}{y,a})}\Ai{bi}
\have{AAbi}{\forall x \forall y (\atom{G}{x,y} \eiff \atom{G}{y,x})}\Ai{Abi}
\end{fitchproof}}
\item $\forall x(\enot \atom{M}{x} \eor \atom{L}{j,x}), \forall x(Bx\eif \atom{L}{j,x}), \forall x(Mx\eor \atom{B}{x})\proves \forall xLjx$
\myanswer{\begin{fitchproof}
\hypo{Anmlj}{\forall x (\enot \atom{M}{x} \eor \atom{L}{j,x})}
\hypo{Ablj}{\forall x (\atom{B}{x} \eif \atom{L}{j,x})}
\hypo{Amb}{\forall x (\atom{M}{x} \eor \atom{B}{x})}
\have{nmlj}{\enot \atom{M}{a} \eor \atom{L}{j,x}}\Ae{Anmlj}
\have{blj}{\atom{B}{a} \eif \atom{L}{j,a}}\Ae{Ablj}
\have{mb}{\atom{M}{a} \eor \atom{B}{a}}\Ae{Amb}
\open
\hypo{nm}{\enot \atom{M}{a}}
\have{b}{\atom{B}{a}}\ds{mb, nm}
\have{lj}{\atom{L}{j,a}}\ce{blj, b}
\close
\open
\hypo{lj1}{\atom{L}{j,a}}
\have{lj2}{\atom{L}{j,a}}\by{R}{lj1}
\close
\have{lj3}{\atom{L}{j,a}}\oe{nmlj, nm-lj, lj1-lj2}
\have{Alj}{\forall x\,\atom{L}{j,x}}\Ai{lj3}
\end{fitchproof}}
\end{earg}
\
\problempart
\label{pr.likes}
Geben Sie einen Symbolisierungsschlüssel für das folgende Argument, symbolisieren Sie es und geben Sie einen entsprechenden Beweis:
\begin{quote}
Es gibt jemanden, der jeden mag, der jeden mag, den er mag. Es gibt also jemanden, der sich selbst mag.
\end{quote}
\myanswer{\begin{ekey}
\item[\text{Domäne}] Personen
\item[M(x,y)] \gap{x} Mag \gap{y}
\end{ekey}
$$\exists x \forall y(\forall z(\atom{M}{x,z} \eif \atom{M}{y,z}) \eif \atom{M}{x,y}) \therefore \exists x Lxx$$
\begin{fitchproof}
\hypo{1}{\exists x\forall y(\forall z(\atom{M}{x,z} \eif \atom{M}{y,z}) \eif \atom{M}{x,y})}
\open
\hypo{a}{\forall y(\forall z(\atom{M}{a,z} \eif \atom{M}{y,z}) \eif \atom{M}{a,y})}
\have{b}{\forall z(\atom{M}{a,z} \eif \atom{M}{a,z}) \eif \atom{M}{a,a}} \Ae{a}
\open
\hypo{lac}{\atom{M}{a,c}}
\have{lac1}{\atom{M}{a,c}}\by{R}{lac}
\close
\have{laclac}{\atom{M}{a,c} \eif \atom{M}{a,c}}\ci{lac-lac1}
\have{Alaz}{\forall z (\atom{M}{a,z} \eif \atom{M}{a,z})}\Ai{laclac}
\have{laa}{\atom{M}{a,a}}\ce{b, Alaz}
\have{El}{\exists x\,\atom{M}{x,x}}\Ei{laa}
\close
\have{n}{\exists x\,\atom{M}{x,x}} \Ee{1, a--El}
\end{fitchproof}}
\
\problempart
Zeigen Sie, dass die folgenden Satzpaare beweisbar äquivalent sind.
\begin{earg}
\newpage
\item $\forall x (\atom{A}{x} \eif \enot \atom{B}{x})$, $\enot\exists x(\atom{A}{x} \eand \atom{B}{x})$
\myanswer{\begin{fitchproof}
\hypo{1}{\forall x (\atom{A}{x} \eif \enot \atom{B}{x})}
\open
\hypo{2}{\exists x(\atom{A}{x} \eand \atom{B}{x})}
\open
\hypo{3}{\atom{A}{c} \eand \atom{B}{c})}
\have{4}{\atom{A}{c}}\ae{3}
\have{5}{\atom{A}{c} \eif \enot \atom{B}{c}}\Ae{1}
\have{6}{\enot \atom{B}{c}}\ce{4, 5}
\have{7}{\atom{B}{c}}\ae{3}
\have{8}{\ered}\ne{6, 7}
\close
\have{9}{\ered}\Ee{2, 3-8}
\close
\have{n}{\enot \exists x(\atom{A}{x} \eand \atom{B}{x})}\ni{2-9}
\end{fitchproof}
\begin{fitchproof}
\hypo{1}{\enot \exists x(\atom{A}{x} \eand \atom{B}{x})}
\open
\hypo{2}{\atom{A}{c}}
\open
\hypo{3}{\atom{B}{c}}
\have{4}{\atom{A}{c} \eand \atom{B}{c}}\ai{2,3}
\have{5}{\exists x(\atom{A}{x} \eand \atom{B}{x})}\Ei{4}
\have{6}{\ered}\ne{1, 5}
\close
\have{7}{\enot \atom{B}{c}}\ni{3-6}
\close
\have{8}{\atom{A}{c} \eif \enot \atom{B}{c}}\ci{2-7}
\have{n}{\forall x (\atom{A}{x} \eif \enot \atom{B}{x})}\Ai{8}
\end{fitchproof}}
\newpage
\item $\forall x (\enot \atom{A}{x}\eif \atom{B}{d})$, $\forall x\,\atom{A}{x} \eor \atom{B}{d}$
\myanswer{\begin{fitchproof}
\hypo{1}{\forall x (\enot \atom{A}{x}\eif \atom{B}{d})}
\open
\hypo{2}{\enot (\forall x\,\atom{A}{x} \eor \atom{B}{d})}
\have{3}{\enot \atom{A}{c} \eif \atom{B}{d}}\Ae{1}
\open
\hypo{4}{\enot \atom{A}{c}}
\have{5}{\atom{B}{d}}\ce{3, 4}
\have{6}{\atom{A}{a} \eor \atom{B}{d}}\oi{5}
\have{7}{\ered}\ne{3, 6}
\close
\have{8}{\atom{A}{c}}\ip{4-7}
\have{9}{\forall x\,\atom{A}{x}}\Ai{8}
\have{10}{\forall x\,\atom{A}{x} \eor \atom{B}{d}}\oi{9}
\have{11}{\ered}\ne{2, 10}
\close
\have{n}{\forall x\,\atom{A}{x} \eor \atom{B}{d}}\ip{2-11}
\end{fitchproof}
\begin{fitchproof}
\hypo{1}{\forall x\,\atom{A}{x} \eor \atom{B}{d}}
\open
\hypo{2}{\enot \atom{A}{c}}
\open
\hypo{3}{\forall x\,\atom{A}{x}}
\have{4}{\atom{A}{c}}\Ae{3}
\have{5}{\ered}\ne{2, 4}
\have{6}{\atom{B}{d}}\re{5}
\close
\open
\hypo{7}{\atom{B}{d}}
\have{8}{\atom{B}{d}}\by{R}{7}
\close
\have{9}{\atom{B}{d}}\oe{1, 3-6, 7-8}
\close
\have{10}{\enot \atom{A}{c} \eif \atom{B}{d}}\ci{2-9}
\have{n}{\forall x (\enot \atom{A}{x}\eif \atom{B}{d})}\Ai{10}
\end{fitchproof}}
\newpage
\item $\exists x\,\atom{P}{x} \eif \atom{Q}{c}$, $\forall x (\atom{P}{x} \eif \atom{Q}{c})$
\myanswer{\begin{fitchproof}
\hypo{1}{\exists x\,\atom{P}{x} \eif \atom{Q}{c}}
\open
\hypo{2}{\atom{P}{a}}
\have{3}{\exists x \atom{P}{x}}\Ei{2}
\have{4}{\atom{Q}{c}}\ce{1, 3}
\close
\have{5}{\atom{P}{a} \eif \atom{Q}{c}}\ci{2-4}
\have{n}{\forall x (\atom{P}{x} \eif \atom{Q}{c}}\Ai{5}
\end{fitchproof}
\begin{fitchproof}
\hypo{1}{\forall x (\atom{P}{x} \eif \atom{Q}{c}}
\open
\hypo{2}{\exists x\,\atom{P}{x}}
\open
\hypo{3}{\atom{P}{a}}
\have{4}{\atom{P}{a} \eif \atom{Q}{c}}\Ae{1}
\have{5}{\atom{Q}{c}}\ce{4, 3}
\close
\have{6}{\atom{Q}{c}}\Ee{2, 3-5}
\close
\have{n}{\exists x\,\atom{P}{x} \eif \atom{Q}{c}}\ci{2-6}
\end{fitchproof}}
\end{earg}
\
\problempart
\label{pr.FOLequivornot}
Für jedes der folgenden Satzpaare: Wenn sie beweisbar äquivalent sind, legen Sie Beweise vor, um dies zu belegen. Wenn sie nicht äquivalent sind, konstruieren Sie eine Interpretation, um zu zeigen, dass sie es nicht sind.
\begin{earg}
\item $\forall x\,\atom{P}{x} \eif \atom{Q}{c}, \forall x (\atom{P}{x} \eif \atom{Q}{c})$
\hfill \myanswer{Nicht äquivalent.
\\ Gegeninterpretation: Die Domäne beinhaltet Zahlen $1$ und $2$. `$c$' benennt $1$. `$Px$' trifft auf und nur auf $1$ zu. `$Qx$' trifft auf und nur auf $2$ zu.}
\item $\forall x\forall y \forall z Bxyz, \forall x\,\atom{B}{x,x}x$
\hfill \myanswer{Nicht äquivalent.
\\ Gegeninterpretation: Domäne beinhaltet $1$ und $2$. `$Bxyz$' trifft auf und nur auf \ntuple{1,1,1} und \ntuple{2,2,2} zu.}
\item $\forall x\forall y\,\atom{D}{x,y}, \forall y\forall x\,\atom{D}{x,y}$
\hfill \myanswer{Beweisbar äquivalent.
\begin{multicols}{2}
\begin{fitchproof}
\hypo{AAd}{\forall x \forall y\,\atom{D}{x,y}}
\have{Ad}{\forall y\,\atom{D}{a,y}}\Ae{AAd}
\have{A}{\atom{D}{a,b}}\Ae{Ad}
\have{Ad1}{\forall x\,\atom{D}{x,b}}\Ai{A}
\have{AAd1}{\forall y \forall x\,\atom{D}{x,y}}\Ai{Ad1}
\end{fitchproof}
\begin{fitchproof}
\hypo{AAd}{\forall y \forall x\,\atom{D}{x,y}}
\have{Ad}{\forall x\,\atom{D}{x,a}}\Ae{AAd}
\have{A}{\atom{D}{b,a}}\Ae{Ad}
\have{Ad1}{\forall y\,\atom{D}{b,y}}\Ai{A}
\have{AAd1}{\forall x \forall y\,\atom{D}{x,y}}\Ai{Ad1}
\end{fitchproof}
\end{multicols}}
\item $\exists x\forall y\,\atom{D}{x,y}, \forall y\exists x\,\atom{D}{x,y}$
\hfill \myanswer{Nicht äquivalent.
\\ Gegeninterpretation: Domäne beinhaltet $1$ und $2$. `$Dxy$' trifft auf und nur auf \ntuple{1,2} und \ntuple{2,1} zu.}
\item $\forall x (\atom{R}{c,a} \eiff \atom{R}{x,a}), Rca \eiff \forall x\,\atom{R}{x,a}$
\hfill \myanswer{Nicht äquivalent.
\\ Diagrammatische Gegeninterpretation, zusätzlich benennt `$a$' 1 and `$c$' 2:
\begin{center}
\begin{tikzpicture}
\node (atom4) at (0,0) {1};
\node (atom5) at (2,0) {2};
\draw[->, thick] (atom4)+(-0.15,0.15) arc (-330:-30:.3);
\end{tikzpicture}
\end{center}}
\end{earg}
\problempart
\label{pr.FOLvalidornot}
Für jedes der folgenden Argumente: Wenn es in der LEO gültig ist, geben Sie einen Beweis. Wenn es ungültig ist, konstruieren Sie eine Interpretation, um zu zeigen, dass es ungültig ist.
\begin{earg}
\item $\exists y\forall x\,\atom{R}{x,y} \therefore \forall x\exists y\,\atom{R}{x,y}$
\hfill \myanswer{Gültig.
\begin{fitchproof}
\hypo{EAr}{\exists y \forall x\,\atom{R}{x,y}}
\open
\hypo{Ar}{\forall x\,\atom{R}{x,a}}
\have{r}{\atom{R}{b,a}}\Ae{Ar}
\have{Er}{\exists y\,\atom{R}{b,y}}\Ei{r}
\close
\have{Er1}{\exists y\,\atom{R}{b,y}}\Ee{EAr, Ar-Er}
\have{AEr}{\forall x \exists y\,\atom{R}{x,y}}\Ai{Er1}
\end{fitchproof}}
\item $\forall x\exists y\,\atom{R}{x,y} \therefore \exists y\forall x\,\atom{R}{x,y}$
\hfill \myanswer{Nicht gültig.
\\ Gegeninterpretation: Domäne beinhaltet $1$ und $2$. `$Rxy$' trifft auf und nur auf \ntuple{1, 2} und \ntuple{2, 1} zu.}
\item $\exists x(\atom{P}{x} \eand \enot \atom{Q}{x}) \therefore \forall x(\atom{P}{x} \eif \enot \atom{Q}{x})$
\hfill \myanswer{Nicht gültig.
\\ Gegeninterpretation: Domäne beinhaltet $1$ und $2$. `$Px$' trifft auf alles in der Domäne zu. `$Qx$' trifft auf und nur auf $2$ zu.}
\item $\forall x(\atom{S}{x} \eif \atom{T}{a}), \atom{S}{d} \therefore \atom{T}{a}$
\hfill \myanswer{Gültig.
\begin{fitchproof}
\hypo{Ast}{\forall x (\atom{S}{x} \eif \atom{T}{a})}
\hypo{s}{\atom{S}{d}}
\have{st}{\atom{S}{d} \eif \atom{T}{a}}\Ae{Ast}
\have{t}{\atom{T}{a}}\ce{st, s}
\end{fitchproof}}
\item $\forall x(Ax\eif \atom{B}{x}), \forall x(\atom{B}{x} \eif \atom{C}{x}) \therefore \forall x(\atom{A}{x} \eif \atom{C}{x})$
\hfill \myanswer{Gültig.
\begin{fitchproof}
\hypo{Aab}{\forall x (\atom{A}{x} \eif \atom{B}{x})}
\hypo{Abc}{\forall x (\atom{B}{x} \eif \atom{C}{x})}
\have{ab}{\atom{A}{a} \eif \atom{B}{a}}\Ae{Aab}
\have{bc}{\atom{B}{a} \eif \atom{C}{a}}\Ae{Abc}
\open
\hypo{a}{\atom{A}{a}}
\have{b}{\atom{B}{a}}\ce{ab, a}
\have{c}{\atom{C}{a}}\ce{bc, b}
\close
\have{ac}{\atom{A}{a} \eif \atom{C}{a}}\ci{a-c}
\have{Aac}{\forall x (\atom{A}{x} \eif \atom{C}{x})}\Ai{ac}
\end{fitchproof}}
\item $\exists x(\atom{D}{x} \eor \atom{E}{x}), \forall x(\atom{D}{x} \eif \atom{F}{x}) \therefore \exists x(\atom{D}{x} \eand \atom{F}{x})$
\hfill \myanswer{Ungültig.
\\ Gegeninterpretation: Domäne beinhaltet $1$. `$Dx$' trifft auf nichts zu. Sowohl `$Ex$' als auch `$Fx$' treffen auf alles in der Domäne zu.}
\item $\forall x\forall y(\atom{R}{x,y} \eor \atom{R}{y,x}) \therefore Rjj$
\hfill \myanswer{Gültig.
\begin{fitchproof}
\hypo{AArr}{\forall x \forall y (\atom{R}{x,y} \eor \atom{R}{y,x})}
\have{Arr}{\forall y (\atom{R}{j,y} \eor \atom{R}{y,j})}\Ae{AArr}
\have{rr}{\atom{R}{j,j} \eor \atom{R}{j,j}}\Ae{Arr}
\open
\hypo{r1}{\atom{R}{j,j}}
\have{r2}{\atom{R}{j,j}}\by{R}{r1}
\close
\open
\hypo{r3}{\atom{R}{j,j}}
\have{r4}{\atom{R}{j,j}}\by{R}{r3}
\close
\have{r5}{\atom{R}{j,j}}\oe{rr, r1-r2, r3-r4}
\end{fitchproof}}
\item $\exists x\exists y(\atom{R}{x,y} \eor \atom{R}{y,x}) \therefore Rjj$
\hfill \myanswer{Ungültig.
\\ Diagrammatische Gegeninterpretation, mit `$j$' als Name für $2$.
\begin{center}
\begin{tikzpicture}
\node (atom4) at (0,0) {1};
\node (atom5) at (2,0) {2};
\draw[->, thick] (atom4)+(-0.15,0.15) arc (-330:-30:.3);
\end{tikzpicture}
\end{center}}
\item $\forall x\,\atom{P}{x} \eif \forall x\,\atom{Q}{x}, \exists x \enot \atom{P}{x} \therefore \exists x \enot \atom{Q}{x}$
\hfill \myanswer{Ungültig.
\\ Gegeninterpretation: Domäne beinhaltet $1$. `$Px$' trifft auf nichts zu. `$Qx$' trifft auf $1$ zu.}
\item $\exists x\,\atom{M}{x} \eif \exists x\,\atom{N}{x}$, $\enot \exists x\,\atom{N}{x}\therefore \forall x \enot \atom{M}{x}$
\hfill \myanswer{Gültig.
\begin{fitchproof}
\hypo{EMiEN}{\exists x\,\atom{M}{x} \eif \exists x\,\atom{N}{x}}
\hypo{nEN}{\enot \exists x\,\atom{N}{x}}
\open
\hypo{Ma}{\atom{M}{a}}
\have{EM}{\exists x\,\atom{M}{x}}\Ei{Ma}
\have{EN}{\exists x\,\atom{N}{x}}\ce{EMiEN,EM}
\have{abs}{\ered}\ri{EN, nEN}
\close
\have{nMa}{\enot \atom{M}{a}}\ni{Ma-abs}
\have{AnM}{\forall x \enot \atom{M}{x}}\Ai{nMa}
\end{fitchproof}}
\end{earg}
\chapter{Beweise mit Quantoren}\setcounter{ProbPart}{0}
\problempart
Verwenden Sie die obigen Strategien, um Beweise für jedes der folgenden Argumente und Theoreme zu finden:
\begin{earg}
\item $A \eif \forall x\,\atom{B}{x} \therefore \forall x(A \eif \atom{B}{x})$
\myanswer{
\begin{fitchproof}
\hypo{1}{A \eif \forall x\,\atom{B}{x}}
\open
\hypo{2}{A}
\have{3}{\forall x\,\atom{B}{x}}\ce{1, 2}
\have{4}{\atom{B}{c}}\Ae{3}
\close
\have{5}{A \eif \atom{B}{c}}\ci{2-4}
\have{n}{\forall x(A \eif \atom{B}{x})}\Ai{5}
\end{fitchproof}}
\item $\exists x(A \eif \atom{B}{x}) \therefore A \eif \exists x\, \atom{B}{x}$
\myanswer{
\begin{fitchproof}
\hypo{1}{\exists x(A \eif \atom{B}{x})}
\open
\hypo{2}{A}
\open
\hypo{3}{A \eif \atom{B}{c}}
\have{4}{\atom{B}{c}}\ce{3, 2}
\have{5}{\exists x\, \atom{B}{x}}\Ei{4}
\close
\have{6}{\exists x\, \atom{B}{x}}\Ee{1, 3-5}
\close
\have{n}{A \eif \exists x\, \atom{B}{x}}\ci{2-6}
\end{fitchproof}}
\newpage
\item $\therefore \forall x(\atom{A}{x} \eand \atom{B}{x}) \eiff (\forall x\,\atom{A}{x} \eand \forall x\,\atom{B}{x})$
\myanswer{
\begin{fitchproof}
\open
\hypo{1}{\forall x(\atom{A}{x} \eand \atom{B}{x})}
\have{2}{\atom{A}{c} \eand \atom{B}{c}}\Ae{1}
\have{3}{\atom{A}{c}}\ae{2}
\have{4}{\forall x \atom{A}{x}}\Ai{3}
\have{5}{\atom{B}{c}}\ae{2}
\have{6}{\forall x \atom{B}{x}}\Ai{5}
\have{7}{\forall x\,\atom{A}{x} \eand \forall x\,\atom{B}{x}}\ai{4, 6}
\close
\open
\hypo{8}{\forall x\,\atom{A}{x} \eand \forall x\,\atom{B}{x}}
\have{9}{\forall x \atom{A}{x}}\ae{8}
\have{10}{\atom{A}{c}}\Ae{9}
\have{11}{\forall x \atom{B}{x}}\ae{8}
\have{12}{\atom{B}{c}}\Ae{11}
\have{13}{\atom{A}{c} \eand \atom{B}{c}}\ai{10, 12}
\have{14}{\forall x(\atom{A}{x} \eand \atom{B}{x})}\Ai{13}
\close
\have{n}{\forall x(\atom{A}{x} \eand \atom{B}{x}) \eiff (\forall x\,\atom{A}{x} \eand \forall x\,\atom{B}{x})}\bi{1-7, 8-14}
\end{fitchproof}}
\newpage
\item $\therefore \exists x(\atom{A}{x} \eor \atom{B}{x}) \eiff (\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x})$
\myanswer{
\begin{fitchproof}
\open
\hypo{1}{\exists x(\atom{A}{x} \eor \atom{B}{x})}
\open
\hypo{2}{\atom{A}{c} \eor \atom{B}{c}}
\open
\hypo{3}{\atom{A}{c}}
\have{4}{\exists x \atom{A}{x}}\Ei{3}
\have{5}{\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x}}\oi{4}
\close
\open
\hypo{6}{\atom{B}{c}}
\have{7}{\exists x \atom{B}{c}}\Ei{6}
\have{8}{\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x}}\oi{7}
\close
\have{9}{\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x}}\oe{2, 3-5, 6-8}
\close
\have{10}{\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x})}\Ee{1, 2-9}
\close
\open
\hypo{11}{\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x}}
\open
\hypo{12}{\exists x\,\atom{A}{x}}
\open
\hypo{13}{\atom{A}{c}}
\have{14}{\atom{A}{c} \eor \atom{B}{c}}\oi{13}
\have{15}{\exists x(\atom{A}{x} \eor \atom{B}{x})}\Ei{14}
\close
\have{16}{\exists x(\atom{A}{x} \eor \atom{B}{x})}\Ee{12, 13-15}
\close
\open
\hypo{17}{\exists x\,\atom{B}{x}}
\open
\hypo{18}{\atom{B}{c}}
\have{19}{\atom{A}{c} \eor \atom{B}{c}}\oi{18}
\have{20}{\exists x(\atom{A}{x} \eor \atom{B}{x})}\Ei{19}
\close
\have{21}{\exists x(\atom{A}{x} \eor \atom{B}{x})}\Ee{17, 18-20}
\close
\have{22}{\exists x(\atom{A}{x} \eor \atom{B}{x})}\oe{11, 12-16, 17-21}
\close
\have{n}{\exists x(\atom{A}{x} \eor \atom{B}{x}) \eiff (\exists x\,\atom{A}{x} \eor \exists x\,\atom{B}{x})}\bi{1-10, 11-22}
\end{fitchproof}}
\newpage
\item $A \eor \forall x\,\atom{B}{x} \therefore \forall x(A \eor \atom{B}{x})$
\myanswer{
\begin{fitchproof}
\hypo{1}{A \eor \forall x\,\atom{B}{x}}
\open
\hypo{2}{A}
\have{3}{A \eor \atom{B}{c}}\oi{2}
\have{4}{\forall x(A \eor \atom{B}{x})}\Ai{3}
\close
\open
\hypo{5}{\forall x\,\atom{B}{x}}
\have{6}{\atom{B}{c}}\Ae{5}
\have{7}{A \eor \atom{B}{c}}\oi{6}
\have{8}{\forall x(A \eor \atom{B}{x})}\Ai{7}
\close
\have{n}{\forall x(A \eor \atom{B}{x})}\oe{2-4, 5-8}
\end{fitchproof}}
\item $\forall x(\atom{A}{x} \eif B) \therefore \exists x\,\atom{A}{x} \eif B$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x(\atom{A}{x} \eif B)}
\open
\hypo{2}{\exists x\,\atom{A}{x}}
\open
\hypo{3}{\atom{A}{c}}
\have{4}{\atom{A}{c} \eif B}\Ae{1}
\have{5}{B}\ce{4, 3}
\close
\have{6}{B}\Ee{2, 3-5}
\close
\have{n}{\exists x\,\atom{A}{x} \eif B}\ci{2-6}
\end{fitchproof}}
\item $\exists x(\atom{A}{x} \eif B) \therefore \forall x\,\atom{A}{x} \eif B$
\myanswer{
\begin{fitchproof}
\hypo{1}{\exists x(\atom{A}{x} \eif B)}
\open
\hypo{2}{\forall x\,\atom{A}{x}}
\open
\hypo{3}{\atom{A}{c} \eif B}
\have{4}{\atom{A}{c}}\Ae{2}
\have{5}{B}\ce{3,4}
\have{6}{B}\Ee{1, 3-5}
\close
\have{n}{\forall x\,\atom{A}{x} \eif B}\ci{2-6}
\end{fitchproof}}
\newpage
\item $\therefore \forall x(\atom{A}{x} \eif \exists y\,\atom{A}{y})$
\myanswer{
\begin{fitchproof}
\open
\hypo{1}{\atom{A}{c}}
\have{2}{\exists y \atom{A}{y}}\Ei{1}
\close
\have{3}{\atom{A}{c} \eif \exists y\,\atom{A}{y}}\ci{1-2}
\have{n}{\forall x(\atom{A}{x} \eif \exists y\,\atom{A}{y})}\Ai{3}
\end{fitchproof}}
\end{earg}
Verwenden Sie zusätzlich zu den grundlegenden Quantorenregeln nur die Grundregeln der WFL.
\
\problempart
Verwenden Sie die obigen Strategien, um Beweise für jedes der folgenden Argumente und Theoreme zu finden:
\begin{earg}
\item $\forall x\,\atom{R}{x,x} \therefore \forall x\,\exists y\,\atom{R}{x,y}$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x\,\atom{R}{x,x}}
\have{2}{\atom{R}{c,c}}\Ae{1}
\have{3}{\exists y \atom{R}{c,y}}\Ei{2}
\have{4}{\forall x \exists y \atom{R}{x,y}}\Ai{3}
\end{fitchproof}}
\item $\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{y,z}) \eif \atom{R}{x,z}]$ \\
$\therefore \forall x\,\forall y[\atom{R}{x,y} \eif \forall z(\atom{R}{y,z} \eif \atom{R}{x,z})]$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{y,z}) \eif \atom{R}{x,z}]}
\have{2}{\forall y\,\forall z[(\atom{R}{a,y} \eand \atom{R}{y,z}) \eif \atom{R}{a,z}]}\Ae{1}
\have{3}{\forall z[(\atom{R}{a,b} \eand \atom{R}{b,z}) \eif \atom{R}{a,z}]}\Ae{2}
\have{4}{(\atom{R}{a,b} \eand \atom{R}{b,c}) \eif \atom{R}{a,c}}
\open
\hypo{5}{\atom{R}{a,b}}
\open
\hypo{6}{\atom{R}{b,c}}
\have{7}{\atom{R}{a,b} \eand \atom{R}{b,c}}\ai{5, 6}
\have{8}{\atom{R}{a,c}}\ce{4, 7}
\close
\have{9}{\atom{R}{b,c} \eif \atom{R}{a,c}}\ci{6-8}
\have{10}{\forall z(\atom{R}{b,z} \eif \atom{R}{a,z})}\Ai{9}
\close
\have{11}{\atom{R}{a,b} \eif \forall z(\atom{R}{b,z} \eif \atom{R}{a,z})}\ci{5-10}
\have{12}{\forall y[\atom{R}{a,y} \eif \forall z(\atom{R}{y,z} \eif \atom{R}{a,z})}\Ai{11}
\have{n}{\forall x\,\forall y[\atom{R}{x,y} \eif \forall z(\atom{R}{y,z} \eif \atom{R}{x,z})]}\Ai{12}
\end{fitchproof}}
\item $\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{y,z}) \eif \atom{R}{x,z}],$\\ $\forall x\,\forall y(\atom{R}{x,y} \eif \atom{R}{y, x})$ \\ $\therefore \forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{x,z}) \eif \atom{R}{y,z}]$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{y,z}) \eif \atom{R}{x,z}]}
\hypo{2}{\forall x\,\forall y(\atom{R}{x,y} \eif \atom{R}{y, x})}
\have{3}{\forall y\,\forall z[(\atom{R}{b,y} \eand \atom{R}{y,z}) \eif \atom{R}{b,z}]}\Ae{1}
\have{4}{\forall z[(\atom{R}{b,a} \eand \atom{R}{a,z}) \eif \atom{R}{b,z}]}\Ae{3}
\have{5}{(\atom{R}{b,a} \eand \atom{R}{a,c}) \eif \atom{R}{b,c}}\Ae{4}
\have{6}{\forall y(\atom{R}{a,y} \eif \atom{R}{y,a})}\Ae{2}
\have{7}{\atom{R}{a,b} \eif \atom{R}{b,a}}\Ae{6}
\open
\hypo{8}{\atom{R}{a,b} \eand \atom{R}{a,c}}
\have{9}{\atom{R}{a,b}}\ae{8}
\have{10}{\atom{R}{b,a}}\ce{7, 9}
\have{11}{\atom{R}{a,c}}\ae{8}
\have{12}{\atom{R}{b,a} \eand \atom{R}{a,c}}\ai{10,11}
\have{13}{\atom{R}{b,c}}\ce{5,12}
\close
\have{14}{(\atom{R}{a,b} \eand \atom{R}{a,c}) \eif \atom{R}{b,c}}\ci{8-13}
\have{15}{\forall z[(\atom{R}{a,b} \eand \atom{R}{a,z}) \eif \atom{R}{b,z}]}\Ai{14}
\have{16}{\forall y\,\forall z[(\atom{R}{a,y} \eand \atom{R}{a,z}) \eif \atom{R}{y,z}]}\Ai{15}
\have{n}{\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{x,z}) \eif \atom{R}{y,z}]}\Ai{16}
\end{fitchproof}}
\newpage
\item $\forall x\,\forall y(\atom{R}{x,y} \eif \atom{R}{y, x})$ \\$\therefore \forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{x,z}) \eif \exists u(\atom{R}{y,u} \eand \atom{R}{z,u})]$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x\,\forall y(\atom{R}{x,y} \eif \atom{R}{y, x})}
\have{2}{\forall y(\atom{R}{a,y} \eif \atom{R}{y,a})}\Ae{1}
\have{3}{\atom{R}{a,b} \eif \atom{R}{b,a}}\Ae{2}
\have{4}{\atom{R}{a,c} \eif \atom{R}{c,a}}\Ae{2}
\open
\hypo{5}{\atom{R}{a,b} \eand \atom{R}{a,c}}
\have{6}{\atom{R}{a,b}}\ae{5}
\have{7}{\atom{R}{b,a}}\ce{3,6}
\have{8}{\atom{R}{a,c}}\ae{5}
\have{9}{\atom{R}{c,a}}\ce{4,8}
\have{10}{\atom{R}{b,a} \eand \atom{R}{c,a}}\ai{7, 9}
\have{11}{\exists u(\atom{R}{b,u} \eand \atom{R}{c,u})}\Ei{10}
\close
\have{12}{(\atom{R}{a,b} \eand \atom{R}{a,c}) \eif \exists u(\atom{R}{b,u} \eand \atom{R}{c,u})}\ci{5-11}
\have{13}{\forall z[(\atom{R}{a,b} \eand \atom{R}{a,z}) \eif \exists u(\atom{R}{b,u} \eand \atom{R}{z,u})]}\Ai{12}
\have{14}{\forall y\,\forall z[(\atom{R}{a,y} \eand \atom{R}{a,z}) \eif \exists u(\atom{R}{y,u} \eand \atom{R}{z,u})]}\Ai{13}
\have{n}{\forall x\,\forall y\,\forall z[(\atom{R}{x,y} \eand \atom{R}{x,z}) \eif \exists u(\atom{R}{y,u} \eand \atom{R}{z,u})]}\Ai{14}
\end{fitchproof}}
\newpage
\item $\therefore \enot \exists x\,\forall y (\atom{A}{x, y} \eiff \lnot\atom{A}{y, y})$
\myanswer{
\begin{fitchproof}
\open
\hypo{1}{\exists x\,\forall y (\atom{A}{x, y} \eiff \enot\atom{A}{y, y})}
\open
\hypo{2}{\forall y (\atom{A}{c, y} \eiff \enot\atom{A}{y, y})}
\have{3}{\atom{A}{c, c} \eiff \enot\atom{A}{c, c}}
\open
\hypo{4}{\atom{A}{c, c}}
\have{5}{\enot \atom{A}{c, c}}\be{3, 4}
\have{6}{\ered}\ne{5, 4}
\close
\have{7}{\enot \atom{A}{c, c}}\ni{4-6}
\have{8}{\atom{A}{c, c}}\be{3, 7}
\have{9}{\ered}\ne{7, 8}
\close
\have{10}{\ered}\Ee{1, 2-9}
\close
\have{n}{\enot \exists x\,\forall y (\atom{A}{x, y} \eiff \enot\atom{A}{y, y})}\ni{1-10}
\end{fitchproof}}
\end{earg}
\
\problempart
Verwenden Sie die obigen Strategien, um Beweise für jedes der folgenden Argumente und Theoreme zu finden:
\begin{earg}
\item $\forall x\,\atom{A}{x} \eif B \therefore \exists x(\atom{A}{x} \eif B)$
\myanswer{
\begin{fitchproof}
\hypo{1}{\forall x \atom{A}{x} \eif B}
\open
\hypo{2}{\enot \exists (\atom{A}{x}\eif B)}
\open
\hypo{3}{\forall x \atom{A}{x}}
\have{4}{B}
\open
\hypo{5}{\atom{A}{c}}
\have{6}{B}\by{R}{4}
\close
\have{7}{\atom{A}{x} \eif B}\ci{5-6}
\have{8}{\exists x (\atom{A}{x} \eif B)}\Ei{7}
\have{9}{\ered}\ne{2, 8}
\close
\have{10}{\enot \forall x \atom{A}{x}}\ni{3-9}
\open
\hypo{11}{\enot \exists x \enot \atom{A}{x}}