\newenvironment처음으로 명령을 사용하는데 문제가 발생합니다. 새로운 theorem환경을 만드는 데 사용하고 있습니다
\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}}
이 새로운 theorem환경을 사용하여 나는 약간의 공간이 필요한 보조정리를 증명하고 있습니다.
\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}
그것은 나를 괴롭히는 증거가 아니므로 아무도 나를 위해 그것을 확인할 필요가 없습니다. ;). 문제는 오히려 증명을 하려면 여러 페이지가 필요하지만 LaTeX는 모든 것을 한 페이지에 담는다는 점이 분명히 조금 불리하다는 점입니다.
따라서 콘텐츠가 여러 페이지에 걸쳐 흐르도록 환경을 조정하는 방법을 누군가가 알려주면 기쁠 것입니다.
답변1
shadethm당신이 보고 있는 동작은 페이지 나누기를 처리할 수 없다는 것을 인정하는 패키지 때문입니다.
패키지 shadethm는 꽤 오래되었습니다. 좋은 소식은 그 이후로 매우 강력하고 사용자 친화적이라는 것입니다.mdframed패키지는 Marco Daniel이 만들었습니다.
이를 사용하면 정말 아름다운 프레임 환경을 만들 수 있으며 페이지 나누기를 처리합니다. 필요에 따라 조정할 수 있는 완전한 MWE가 이어집니다. 자세한 내용은 설명서를 참조하세요.
\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}
기타 참고사항:
$$...$$표시되는 수학 콘텐츠에 을(를) 사용하고 계신 것으로 확인되었습니다 . 이것은 오래된 것이므로 사용해야 합니다\[...\].$$ ... $$보다 \[ ... \]를 선호하는 이유는 무엇입니까?좋은 토론을 위해.- 나는 또한 꽤 많은 수학 환경이 연속적으로 표시되어 있다는 것을 알았습니다. 이 또한 피해야 하며 아마도
amsmath다음과 같은 패키지 의 환경 중 하나를 사용해야 합니다gather.