Si intento meter un árbol forestal grande en una página ampliándola en 3 líneas, obtengo el siguiente resultado. El árbol se superpone con el pie de página.
A continuación se muestra el código. ¿Existe una manera sencilla de meterlo en una página sin permitir que se superponga con el pie de página? ¿O hay alguna manera de dividirlo en dos páginas sin los trucos complejos que me presentaron anteriormente?
\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}
Respuesta1
Podrías usarlo \smallpara hacer las cosas más pequeñas (pero más difíciles de leer :-) o usar \thspagestyle{empty}para perder el número de página, pero suponiendo que no quieras hacer eso, puedes jugar con el espacio en blanco para que las cosas encajen (casi).
aquí utilizo \enlargethispage*en lugar de \enlaregethispagehacerlo, el espacio en blanco se reduce tanto como sea posible y luego ajusto el espacio en blanco alrededor de las enumeraciones usando las enumitemopciones.
\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}