Estoy usando el \newenvironmentcomando por primera vez y está causando problemas. Lo estoy usando para crear un nuevo theorementorno.
\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{amsthm , amsfonts, latexsym}
\usepackage{tikz}
\usepackage{shadethm}
\usepackage[mathscr]{euscript}
\usepackage{graphics,graphicx}
\usepackage{enumerate}
\usepackage{color}
\theoremstyle{definition}
\newshadetheorem{lems}{Lemma}[section]
\newenvironment{lem}[1][]{
\definecolor{shadethmcolor}{HTML}{00FFFF}
\begin{lems}[#1]\hspace*{1mm}
}{\end{lems}}
Usando este nuevo theorementorno estoy haciendo una prueba de un lema que necesita algunas páginas de espacio.
\begin{document}
\begin{lem}
Let $H$ be a connected S-thin hypergraph, and $A \boxtimes B$ and $C \boxtimes D$ be two
decompositions of $H$ with respect to the strong product. Then there exists a decomposition
$$ A_C \boxtimes A_D \boxtimes B_C \boxtimes B_D $$
of $H$ such that $A = A_C \boxtimes A_D, B= B_C \boxtimes B_D , C= A_C \boxtimes B_C,
D=A_D \boxtimes B_D$.
\begin{proof}
The idea of the proof is by using the PFD of the cartesian skeleton $H^{\Box}$ of $H$ to define four proper factors, such that
each factor has as vertex set the set of one of our desired factors, e.g. $A_C$. Then we define the desired factors of $H$ by defining projections on $H$ where the vertex set is given by the composed factors of the cartesian skeleton and such that the
edges obey the definition of the strong product. Finally via those projections it is shown that $A=A_C \boxtimes A_D$.\
Let $ H_1 \Box H_2 \Box ... \Box H_n $ be the unique PFD of a cartesian skeleton $H^{\Box}$ of $H$ . Let $I_A$ be the subset of the index set $ {1,2,...,n}$ with $V(A)=V(\Box_{i \in I_A} H_i)$
and $I_B, I_C$ and $I_D$ be defined analogously. Furthermore set
$$ H_{A,C} = \Box_{i \in I_A \cap I_C} H_i$$
and define $H_{A,D}, H_{B,C}$ and $H_{B,D}$ similarly. Then
$$ H^{\Box} = H_{A,C} \Box H_{A,D} \Box H_{B,C} \Box H_{B,D} . $$
It will be convenient to use only four coordinates $(x_1,x_2,x_3,x_4)$ for every vertex $x \in V(G)$ henceforth. Of course it is possible that not all of the intersections $I_A \cap I_C, I_A \cap I_D, I_D \cap I_C$
and $I_B \cap I_D$ are nonempty. Suppose that $I_B \cap I_C = \emptyset$ then $I_A \cap I_D \neq \emptyset$. If in addition $I_A \cap I_C$ were empty, then $I_A = I_D$ and thus $I_B=I_C$, but then
would be nothing to prove. \
We can thus assume that all but possibly $I_B \cap I_D$ are nonempty and at least three of the four
coordinates are nontrivial, that is to say, there are at least two vertices that differ in the first, second
and third coordinates, but it is possible that all vertices have the same fourth coordinate. \
Clearly, for $y=(y_1,y_2,y_3,y_4),$
$$V(A^y) = { (x_1,x_2,y_3,y_4) | x_1 \in V(H_{A,C}), x_2 \in V(H_{A,D}) } $$
$$V(B^y) = { (y_1,y_2,x_3,x_4) | x_3 \in V(H_{B,C}), x_4 \in V(H_{B,D}) } $$
$$V(C^y) = { (x_1,y_2,x_3,y_4) | x_1 \in V(H_{A,C}), x_3 \in V(H_{B,C}) } $$
$$V(D^y) = { (y_1,x_2,y_3,x_4) | x_2 \in V(H_{A,D}), x_4 \in V(H_{B,D}) } $$
are the vertex sets of the $A-,B-,C-$ and $D-$ layers of $H$. \
We now define $A_C$ as $p_1(H)$, namely $V(A_C) = V(H_{A,C})$ and $${x_1^1,...,x_k^1 } \in E(A_C)$$
if and only if there are vertices $ \tilde x_1=(x_1^1,x_1^2,x_1^3,x_1^4),...,
\tilde x_k =(x_k^1,x_k^2,x_k^3,x_k^4) \in H \text { such that} $
$$ \exists e \in E(H) : \exists S \subseteq I={2,3,4} : p_1(e) = { x_1^1,...,x_k^1} $$
\begin{itemize}
\item[(i)] $ p_s(e) \subseteq e_s \in E_s , \forall s \in S$
\item[(ii)]$ |e| = |p_s(e)|, \forall s \in S$
\item[(iii)] $|p_i(e)|=1, \forall i \in I \setminus S$
\end{itemize}
It is clear what is meant by $A_D, B_C$ and $B_D$. For the proof of the lemma it suffices to show that
$A = A_C \boxtimes A_D$. Recall that $A$ is obtained by projection of $H$ onto the vertex set of $A$. We
call this projection $p_A$ and define $p_B,p_C,p_D$ analogously. With our present coordinatization we thus have
$$p_A(x_1,x_2,x_3,x_4) = (x_1,x_2,-,-), $$
$$p_B(x_1,x_2,x_3,x_4) = (-,-,x_3,x_4), $$
$$p_C(x_1,x_2,x_3,x_4) = (x_1,-,x_3,-), $$
$$p_D(x_1,x_2,x_3,x_4) = (-,x_2,-,x_4). $$
In order to show that $A=A_C \boxtimes A_D$, it suffices to prove that for $e={\tilde x_1,..., \tilde x_k } \in H$ holds
$$ p_A(e) \in A $$ if and only if
$$ p_1(e) \in A_C \text{ and either is } e \text{ an edge in }A_D \text{, hence } p_2(e) \subseteq e \in A_D, |e|=|p_2(e)| $$
$$\text{ or not and thus } |p_2(e)|=1 $$
(wlog. we assume that the edge of $A_C$ is of minimal rank, otherwise we could easily add a case to the above definition
where the edge of $A_D$ was the minimal one. But for the sake of clarity we omit this case).
Suppose that $ {p_A \tilde x_1,..., p_A \tilde x_k } \subseteq e \in A$. We can assume, wlog., that
$\tilde x_1, ... , \tilde x_k $ are chosen such that ${ \tilde x_1 ,... , \tilde x_k } \in E(H).$ But then
$ {p_1 \tilde x_1, ... , p_1 \tilde x_k } \in A_C $ and ${p_2 \tilde x_1, ... , p_2 \tilde x_k } \in A_D$
by the definition of $A_C$ and $A_D$. \
On the other hand, suppose that the edge
${ (x_1^1,-,-,-),(x_2^1,-,-,-),...,(x_k^1,-,-,-}$ is in $A_C$ and ${(_,x_1^2,-,-),...,(-,x_t^2,-,-)} $ is
in $A_D , t \leq t $. \
Then there are vertices $\tilde x_1, \tilde x_2, ... , \tilde x_k, ... , \tilde x_t$ of the form
$$ \tilde x_i = ( x_i^1, x_i^2, x_i^3, x_i^4) $$
with ${ \tilde x_1 , ... , \tilde x_k } \in E(H)$ and ${ \tilde x_1 , ... ,\tilde x_t } \in E(H)$. \
From this we infer that there are edges
$$ { (x_1^1,-,x_1^3,-),...,(x_s^1,-,x_s^3,-) } \in E(C) , k \leq s \leq t $$
$$ {(-,x_1^2,-,x_1^4),...,(-,x_r^2,-,x_r^4) } \in E(D), k \leq r \leq t $$
Since $H = C \boxtimes D$ this implies that
$$ {(x_1^1,x_1^2,x_1^3,x_1^4),...,(x_s^1,x_s^2,x_s^3,x_s^4) } \in E(H) \text{ where wlog. } s \leq r$$
and hence ${(x_1^1,x_1^2,-,-),...,(x_s^1,x_s^2,-,-) } \subseteq e \in E(A)$.
\end{proof}
\end{lem}
\end{document}
No es la prueba lo que me preocupa, por lo tanto nadie tiene que comprobarla por mí;). El problema es más bien que la prueba necesita varias páginas pero LaTeX lo pone todo en una sola, lo que obviamente es un poco desventajoso.
Por lo tanto, me alegraría que alguien pudiera decirme cómo ajustar el entorno de tal manera que el contenido fluya entre las páginas.
Respuesta1
El comportamiento que estás viendo se debe al shadethmpaquete, que admite que no puede manejar saltos de página.
El shadethmpaquete es bastante antiguo; la buena noticia es que desde entonces el extremadamente potente y fácil de usarmdframedEl paquete ha sido creado por Marco Daniel.
Puede crear entornos enmarcados verdaderamente hermosos usándolo y maneja saltos de página. A continuación aparece un MWE completo, que puede modificar según sea necesario; consulte la documentación para obtener más detalles.
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{tikz}
\usepackage[framemethod=tikz]{mdframed}
\usepackage{lipsum}
\newmdtheoremenv[outerlinewidth=3,
innerlinewidth=2,linecolor=gray,
backgroundcolor=blue!20,%
innerlinecolor=blue!50,outerlinecolor=red!50,innertopmargin=0pt,%
splittopskip=\topskip,skipbelow=0pt,%
]{lem}{Lemma}[section]
\begin{document}
\begin{lem}
Let $H$ be a connected S-thin hypergraph, and $A \boxtimes B$ and $C \boxtimes D$ be two
decompositions of $H$ with respect to the strong product. Then there exists a decomposition
$$ A_C \boxtimes A_D \boxtimes B_C \boxtimes B_D $$
of $H$ such that $A = A_C \boxtimes A_D, B= B_C \boxtimes B_D , C= A_C \boxtimes B_C,
D=A_D \boxtimes B_D$.
\mbox{} % needed because you end your lemma with mathematical content
\begin{proof}
\lipsum
\end{proof}
\end{lem}
\end{document}
Algunas otras notas:
- Noté que estabas usando
$$...$$para tu contenido matemático mostrado. Esto está desactualizado y deberías usarlo\[...\], consulta¿Por qué es preferible \[ ... \] a $$ ... $$?para una buena discusión. - También noté que tenía bastantes entornos matemáticos mostrados uno tras otro. Esto también debería evitarse y debería utilizar uno de los entornos del
amsmathpaquete, algo así comogatherquizás.