Wenn ich versuche, einen großen Waldbaum auf eine Seite zu quetschen, indem ich die Seite um drei Zeilen vergrößere, erhalte ich das folgende Ergebnis. Der Baum überlappt sich mit der Fußzeile.
Unten ist der Code. Gibt es eine einfache Möglichkeit, ihn auf eine Seite zu quetschen, ohne dass er sich mit der Fußzeile überschneidet? Oder gibt es eine Möglichkeit, ihn auf zwei Seiten aufzuteilen, ohne die komplexen Tricks, die ich zuvor kennengelernt habe?
\documentclass[oneside,12pt]{article}
\usepackage{geometry}
\usepackage{microtype}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{enumitem}
\setlist[enumerate,1]{label=\bfseries\Alph*,align=left,leftmargin=*,
labelsep=1.5em}
\setlist[enumerate,2]{label=\arabic*.,labelindent=1em,labelsep=1.5em,
leftmargin=*}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\theoremstyle{definition}
\newtheorem*{sol}{Solution}
\usepackage{forest}
\forestset{
ass/.append style={
before computing xy={l=\baselineskip},
no edge
},
prooftree/.style={
baseline,
for tree={
child anchor=north,
parent anchor=south
}
}
}
\begin{document}
\subsection*{Exercises 35}
\begin{enumerate}
\item Use \textbf{QL\textsuperscript{=}} trees to show the following inferences are valid:
\begin{enumerate}
\item Jack is Fingers. Fingers is never caught. Whoever is never caught
escapes justice. So Jack escapes justice.
\begin{sol}
\begin{forest}
prooftree
[{$j=f$}
[$\neg Cf$,ass
[$\forall x(\neg Cx\supset Ex)$,ass
[$\neg Ej$,ass
[$(\neg Cf\supset Ef)$,ass
[$\neg\neg Cf$ [*,ass]] [$Ef$
[$Ej$,ass [*,ass]]]]]]]]
\end{forest}
\end{sol}
\item There is a wise philosopher. There is a philosopher who isn’t wise.
So there are at least two philosophers.
\begin{sol}
\begin{forest}
prooftree
[$\exists x(Fx\wedge Gx)\checkmark$
[$\exists x(Fx\wedge\neg Gx)\checkmark$,ass
[{$\neg\exists x\exists y((Fx\wedge Fy)\wedge\neg x=y)\checkmark$},ass
[{$\forall x\neg\exists y((Fx\wedge Fy)\wedge\neg x=y)$},ass
[$(Fa\wedge Ga)\checkmark$,ass
[$(Fb\wedge\neg Gb)\checkmark$,ass
[$Fa$,ass
[$Ga$,ass
[$Fb$,ass
[$\neg Gb$,ass
[{$\neg\exists y((Fa\wedge Fy)\wedge\neg a=y)\checkmark$},ass
[{$\forall y\neg((Fa\wedge Fy)\wedge\neg a=y)$},ass
[{$\neg((Fa\wedge Fb)\wedge\neg a=b)\checkmark$},ass
[$\neg(Fa\wedge Fb)$
[$\neg Fa$ [*,ass]] [$\neg Fb$ [*,ass]]]
[{$\neg\neg a=b$}
[{$a=b$},ass
[$\neg Ga$,ass [*,ass]]]]]]]]]]]]]]]]]
\end{forest}
\end{sol}
\pagebreak
\item Whoever stole the goods, knew the safe combination. Someone
stole the goods, and only Jack knew the safe combination. Hence
Jack stole the goods.
\begin{sol}
\begin{forest}
prooftree
[$\forall x(Sx\supset Cx)$
[$\exists xSx\checkmark$,ass
[$\forall x(Cx\equiv x{=}j)$,ass
[$\neg Sj$,ass
[$Sa$,ass
[$(Sa\supset Ca)\checkmark$,ass
[$\neg Sa$ [*,ass]] [$Ca$
[$(Ca\equiv a{=}j)\checkmark$,ass
[$(Ca\supset a{=}j)\checkmark$,ass
[$(a{=}j\supset Ca)$,ass
[$\neg Ca$ [*,ass]] [$a{=}j$
[$\neg Sa$,ass [*,ass]]]]]]]]]]]]]
\end{forest}
\end{sol}
\item For every number, there’s a larger one. No number is larger than
itself. So for every number, there’s a distinct number which is
larger than it.
\begin{sol}
\begin{forest}
prooftree
[$\forall x\exists y(Lyx)$
[$\forall x\neg Lxx$,ass
[$\neg\forall x\exists y(Lyx\wedge\neg y{=}x)\checkmark$,ass
[$\exists x\neg\exists y(Lyx\wedge\neg y{=}x)\checkmark$,ass
[$\neg\exists y(Lya\wedge\neg y{=}a)\checkmark$,ass
[$\forall y\neg(Lya\wedge\neg y{=}a)$,ass
[$\exists yLya\checkmark$,ass
[$Lba$,ass
[$\neg(Lba\wedge\neg b{=}a)\checkmark$,ass
[$\neg Lba$ [*,ass]] [$\neg\neg b{=}a\checkmark$
[$b{=}a$,ass
[$Laa$,ass
[$\neg Laa$,ass [*,ass]]]]]]]]]]]]]]
\end{forest}
\end{sol}
\end{enumerate}
\pagebreak
\item Show that the following wffs are q-logical truths
\begin{enumerate}
\item $\forall x\forall y(x{=}y\supset(Fx\supset Fy))$
\item $\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))$
\end{enumerate}
Thinking about the structure of these proofs, conclude that each way of filling out
the following schema from §33.1 does indeed yield a q-logical truth:
\begin{enumerate}
\item[(LS)] $\forall v\forall w(v=w\supset(C(\ldots v\ldots v\ldots)\supset
C(\ldots w\ldots w\ldots)))$
\end{enumerate}
\begin{sol}
\leavevmode
\begin{enumerate}
\item $\forall x\forall y(x{=}y\supset(Fx\supset Fy))$
\begin{forest}
prooftree
[$\neg\forall x\forall y(x{=}y\supset(Fx\supset Fy))\checkmark$
[$\exists x\neg\forall y(x{=}y\supset(Fx\supset Fy))\checkmark$,ass
[$\neg\forall y(a{=}y\supset(Fa\supset Fy))\checkmark$,ass
[$\exists y\neg(a{=}y\supset(Fa\supset Fy))\checkmark$,ass
[$\neg(a{=}b\supset(Fa\supset Fb))$,ass
[$a{=}b$,ass
[$\neg(Fa\supset Fb)\checkmark$,ass
[$Fa$,ass
[$\neg Fb$,ass
[$Fb$,ass [*,ass]]]]]]]]]]]
\end{forest}
\enlargethispage{4\baselineskip}
\item $\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))$
\begin{forest}
prooftree
[$\neg\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))\checkmark$
[$\exists y\neg\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\neg\forall z(a{=}z\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\exists z\neg(a{=}z\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\neg(a{=}b\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxb\wedge Fb)))\checkmark$,ass
[$a{=}b$,ass
[$\neg(\forall x(Lxa\wedge Fa)\supset\forall x(Lxb\wedge Fb))\checkmark$,ass
[$\forall x(Lxa\wedge Fa)$,ass
[$\neg\forall x(Lxb\wedge Fb)\checkmark$,ass
[$\exists x\neg(Lxb\wedge Fb)\checkmark$,ass
[$\neg(Lcb\wedge Fb)$,ass
[$(Lca\wedge Fa)\checkmark$,ass
[$Lca$,ass
[$Fa$,ass
[$\neg Lcb$
[$Lcb$,ass [*,ass]]]
[$\neg Fb$
[$Fb$,ass [*,ass]]]]]]]]]]]]]]]]]
\end{forest}
\end{enumerate}
\end{sol}
\end{enumerate}
\end{document}
Antwort1
Sie könnten verwenden, \smallum die Dinge kleiner zu machen (aber schwerer lesbar :-) oder verwenden, \thspagestyle{empty}um die Seitenzahl wegzulassen, aber vorausgesetzt, Sie möchten das nicht, können Sie mit dem Leerraum spielen, damit die Dinge (fast) passen.
hier verwende ich \enlargethispage*eher als \enlaregethispage, damit der Leerraum möglichst verkleinert wird, und passe dann den Leerraum um die Aufzählungen mithilfe der enumitemOptionen an.
\documentclass[oneside,12pt]{article}
\usepackage{geometry}
\usepackage{microtype}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{enumitem}
\setlist[enumerate,1]{label=\bfseries\Alph*,align=left,leftmargin=*,
labelsep=1.5em}
\setlist[enumerate,2]{label=\arabic*.,labelindent=1em,labelsep=1.5em,
leftmargin=*}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\theoremstyle{definition}
\newtheorem*{sol}{Solution}
\usepackage{forest}
\forestset{
ass/.append style={
before computing xy={l=\baselineskip},
no edge
},
prooftree/.style={
baseline,
for tree={
child anchor=north,
parent anchor=south
}
}
}
\begin{document}
\subsection*{Exercises 35}
\begin{enumerate}
\item Use \textbf{QL\textsuperscript{=}} trees to show the following inferences are valid:
\begin{enumerate}
\item Jack is Fingers. Fingers is never caught. Whoever is never caught
escapes justice. So Jack escapes justice.
\begin{sol}
\begin{forest}
prooftree
[{$j=f$}
[$\neg Cf$,ass
[$\forall x(\neg Cx\supset Ex)$,ass
[$\neg Ej$,ass
[$(\neg Cf\supset Ef)$,ass
[$\neg\neg Cf$ [*,ass]] [$Ef$
[$Ej$,ass [*,ass]]]]]]]]
\end{forest}
\end{sol}
\item There is a wise philosopher. There is a philosopher who isn’t wise.
So there are at least two philosophers.
\begin{sol}
\begin{forest}
prooftree
[$\exists x(Fx\wedge Gx)\checkmark$
[$\exists x(Fx\wedge\neg Gx)\checkmark$,ass
[{$\neg\exists x\exists y((Fx\wedge Fy)\wedge\neg x=y)\checkmark$},ass
[{$\forall x\neg\exists y((Fx\wedge Fy)\wedge\neg x=y)$},ass
[$(Fa\wedge Ga)\checkmark$,ass
[$(Fb\wedge\neg Gb)\checkmark$,ass
[$Fa$,ass
[$Ga$,ass
[$Fb$,ass
[$\neg Gb$,ass
[{$\neg\exists y((Fa\wedge Fy)\wedge\neg a=y)\checkmark$},ass
[{$\forall y\neg((Fa\wedge Fy)\wedge\neg a=y)$},ass
[{$\neg((Fa\wedge Fb)\wedge\neg a=b)\checkmark$},ass
[$\neg(Fa\wedge Fb)$
[$\neg Fa$ [*,ass]] [$\neg Fb$ [*,ass]]]
[{$\neg\neg a=b$}
[{$a=b$},ass
[$\neg Ga$,ass [*,ass]]]]]]]]]]]]]]]]]
\end{forest}
\end{sol}
\pagebreak
\item Whoever stole the goods, knew the safe combination. Someone
stole the goods, and only Jack knew the safe combination. Hence
Jack stole the goods.
\begin{sol}
\begin{forest}
prooftree
[$\forall x(Sx\supset Cx)$
[$\exists xSx\checkmark$,ass
[$\forall x(Cx\equiv x{=}j)$,ass
[$\neg Sj$,ass
[$Sa$,ass
[$(Sa\supset Ca)\checkmark$,ass
[$\neg Sa$ [*,ass]] [$Ca$
[$(Ca\equiv a{=}j)\checkmark$,ass
[$(Ca\supset a{=}j)\checkmark$,ass
[$(a{=}j\supset Ca)$,ass
[$\neg Ca$ [*,ass]] [$a{=}j$
[$\neg Sa$,ass [*,ass]]]]]]]]]]]]]
\end{forest}
\end{sol}
\item For every number, there’s a larger one. No number is larger than
itself. So for every number, there’s a distinct number which is
larger than it.
\begin{sol}
\begin{forest}
prooftree
[$\forall x\exists y(Lyx)$
[$\forall x\neg Lxx$,ass
[$\neg\forall x\exists y(Lyx\wedge\neg y{=}x)\checkmark$,ass
[$\exists x\neg\exists y(Lyx\wedge\neg y{=}x)\checkmark$,ass
[$\neg\exists y(Lya\wedge\neg y{=}a)\checkmark$,ass
[$\forall y\neg(Lya\wedge\neg y{=}a)$,ass
[$\exists yLya\checkmark$,ass
[$Lba$,ass
[$\neg(Lba\wedge\neg b{=}a)\checkmark$,ass
[$\neg Lba$ [*,ass]] [$\neg\neg b{=}a\checkmark$
[$b{=}a$,ass
[$Laa$,ass
[$\neg Laa$,ass [*,ass]]]]]]]]]]]]]]
\end{forest}
\end{sol}
\end{enumerate}
\pagebreak
\item Show that the following wffs are q-logical truths
\begin{enumerate}[topsep=0pt]
\item $\forall x\forall y(x{=}y\supset(Fx\supset Fy))$
\item $\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))$
\end{enumerate}
Thinking about the structure of these proofs, conclude that each way of filling out
the following schema from §33.1 does indeed yield a q-logical truth:
\begin{enumerate}[topsep=0pt]
\item[(LS)] $\forall v\forall w(v=w\supset(C(\ldots v\ldots v\ldots)\supset
C(\ldots w\ldots w\ldots)))$
\end{enumerate}
\begin{sol}
\leavevmode
\begin{enumerate}[topsep=0pt,itemindent=-75pt]
\item $\forall x\forall y(x{=}y\supset(Fx\supset Fy))$
\begin{forest}
prooftree
[$\neg\forall x\forall y(x{=}y\supset(Fx\supset Fy))\checkmark$
[$\exists x\neg\forall y(x{=}y\supset(Fx\supset Fy))\checkmark$,ass
[$\neg\forall y(a{=}y\supset(Fa\supset Fy))\checkmark$,ass
[$\exists y\neg(a{=}y\supset(Fa\supset Fy))\checkmark$,ass
[$\neg(a{=}b\supset(Fa\supset Fb))$,ass
[$a{=}b$,ass
[$\neg(Fa\supset Fb)\checkmark$,ass
[$Fa$,ass
[$\neg Fb$,ass
[$Fb$,ass [*,ass]]]]]]]]]]]
\end{forest}
\enlargethispage*{4\baselineskip}
\item $\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))$
\begin{forest}
prooftree
[$\neg\forall y\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))\checkmark$
[$\exists y\neg\forall z(y{=}z\supset(\forall x(Lxy\wedge Fy)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\neg\forall z(a{=}z\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\exists z\neg(a{=}z\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxz\wedge Fz)))\checkmark$,ass
[$\neg(a{=}b\supset(\forall x(Lxa\wedge Fa)\supset
\forall x(Lxb\wedge Fb)))\checkmark$,ass
[$a{=}b$,ass
[$\neg(\forall x(Lxa\wedge Fa)\supset\forall x(Lxb\wedge Fb))\checkmark$,ass
[$\forall x(Lxa\wedge Fa)$,ass
[$\neg\forall x(Lxb\wedge Fb)\checkmark$,ass
[$\exists x\neg(Lxb\wedge Fb)\checkmark$,ass
[$\neg(Lcb\wedge Fb)$,ass
[$(Lca\wedge Fa)\checkmark$,ass
[$Lca$,ass
[$Fa$,ass
[$\neg Lcb$
[$Lcb$,ass [*,ass]]]
[$\neg Fb$
[$Fb$,ass [*,ass]]]]]]]]]]]]]]]]]
\end{forest}
\end{enumerate}
\end{sol}
\end{enumerate}
\end{document}