Latex用mdframed宏包制作精致的定理证明环境

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注本文非原创，来自：http://www.latexstudio.net/archives/4482

mdframed宏包可以轻松搞定一个带框的段落样式，而且可以自动跨页。另外，宏包还有很强的定制设计部分，之前，我们分享的一个定理环境就是用的该宏包进行制作的。本文将比较详细的介绍该宏包的使用。

在导言区，引入mdframed宏包且可以直接定义定理，引理，证明，推论这些环境。引入该宏包的时候，推荐选择TikZ的方法来进行框架绘制。

1. \usepackage[framemethod=TikZ]{mdframed}

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接下来，我们可以定义一个带计数器的环境了，比如，我们比较熟悉的定理环境，我们通常希望这些定理环境能够和小节的序号关联起来，并且能够在小节标题序号变化的时候能够重置。那么我们可以做如下：

1. \newcounter{theo}[section]\setcounter{theo}{0}
   
2. \renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}

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按照一般的环境定义，我们可以定义如下环境：

1. \newenvironment{theo}[2][]{%
   
2.     \refstepcounter{theo}
   
3. 
   
4.     % Code for box design goes here.
   
5. 
   
6. \begin{mdframed}[]\relax}{%
   
7. \end{mdframed}}

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用mdframed设计的定理，计数器的部分是放在一个box里的，因此有时box会有显示定理的具体名称，有时只是单纯的定理序号，因此我们还需要判断不同情况的box设计。我们可以用\ifstringempty{#1}来判断定理里面是否有内容。我们的代码可以改写为：

1. \newcounter{lem}[section]\setcounter{lem}{0}
   
2. \renewcommand{\thelem}{\arabic{section}.\arabic{lem}}
   
3. \newenvironment{lem}[2][]{%
   
4. \refstepcounter{lem}%
   
5. \ifstrempty{#1}%
   
6. {\mdfsetup{%
   
7. frametitle={%
   
8. \tikz[baseline=(current bounding box.east),outer sep=0pt]
   
9. \node[anchor=east,rectangle,fill=green!20]
   
10. {\strut Lemma~\thelem};}}
    
11. }%
    
12. {\mdfsetup{%
    
13. frametitle={%
    
14. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
15. \node[anchor=east,rectangle,fill=green!20]
    
16. {\strut Lemma~\thetheo:~#1};}}%
    
17. }%
    
18. \mdfsetup{innertopmargin=10pt,linecolor=green!20,%
    
19. linewidth=2pt,topline=true,%
    
20. frametitleaboveskip=\dimexpr-\ht\strutbox\relax
    
21. }
    
22. \begin{mdframed}[]\relax%
    
23. \label{#2}}{\end{mdframed}}

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同样的方法，我们可以设计证明环境的样式如下：

1. \newcounter{prf}[section]\setcounter{prf}{0}
   
2. \renewcommand{\theprf}{\arabic{section}.\arabic{prf}}
   
3. \newenvironment{prf}[2][]{%
   
4. \refstepcounter{prf}%
   
5. \ifstrempty{#1}%
   
6. {\mdfsetup{%
   
7. frametitle={%
   
8. \tikz[baseline=(current bounding box.east),outer sep=0pt]
   
9. \node[anchor=east,rectangle,fill=red!20]
   
10. {\strut Proof~\theprf};}}
    
11. }%
    
12. {\mdfsetup{%
    
13. frametitle={%
    
14. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
15. \node[anchor=east,rectangle,fill=red!20]
    
16. {\strut Proof~\thetheo:~#1};}}%
    
17. }%
    
18. \mdfsetup{innertopmargin=10pt,linecolor=red!20,%
    
19. linewidth=2pt,topline=true,%
    
20. frametitleaboveskip=\dimexpr-\ht\strutbox\relax
    
21. }
    
22. \begin{mdframed}[]\relax%
    
23. \label{#2}}{\end{mdframed}}

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通常证明结束的地方，有个结束符，那么我们可以载入amsthm宏包，定义如下：

1. % Load package
   
2. \usepackage{amsthm}
   
3. 
   
4. % Change last line of above code
   
5. \begin{mdframed}[]\relax%
   
6. \label{#2}}{\qed\end{mdframed}}

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完整的代码和效果如下：

1. \documentclass{article}
   
2. \usepackage[framemethod=TikZ]{mdframed}
   
3. \usepackage{amsthm}
   
4. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
   
5. %Theorem
   
6. \newcounter{theo}[section] \setcounter{theo}{0}
   
7. \renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}
   
8. \newenvironment{theo}[2][]{%
   
9. \refstepcounter{theo}%
   
10. \ifstrempty{#1}%
    
11. {\mdfsetup{%
    
12. frametitle={%
    
13. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
14. \node[anchor=east,rectangle,fill=blue!20]
    
15. {\strut Theorem~\thetheo};}}
    
16. }%
    
17. {\mdfsetup{%
    
18. frametitle={%
    
19. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
20. \node[anchor=east,rectangle,fill=blue!20]
    
21. {\strut Theorem~\thetheo:~#1};}}%
    
22. }%
    
23. \mdfsetup{innertopmargin=10pt,linecolor=blue!20,%
    
24. linewidth=2pt,topline=true,%
    
25. frametitleaboveskip=\dimexpr-\ht\strutbox\relax
    
26. }
    
27. \begin{mdframed}[]\relax%
    
28. \label{#2}}{\end{mdframed}}
    
29. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    
30. %Lemma
    
31. \newcounter{lem}[section] \setcounter{lem}{0}
    
32. \renewcommand{\thelem}{\arabic{section}.\arabic{lem}}
    
33. \newenvironment{lem}[2][]{%
    
34. \refstepcounter{lem}%
    
35. \ifstrempty{#1}%
    
36. {\mdfsetup{%
    
37. frametitle={%
    
38. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
39. \node[anchor=east,rectangle,fill=green!20]
    
40. {\strut Lemma~\thelem};}}
    
41. }%
    
42. {\mdfsetup{%
    
43. frametitle={%
    
44. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
45. \node[anchor=east,rectangle,fill=green!20]
    
46. {\strut Lemma~\thetheo:~#1};}}%
    
47. }%
    
48. \mdfsetup{innertopmargin=10pt,linecolor=green!20,%
    
49. linewidth=2pt,topline=true,%
    
50. frametitleaboveskip=\dimexpr-\ht\strutbox\relax
    
51. }
    
52. \begin{mdframed}[]\relax%
    
53. \label{#2}}{\end{mdframed}}
    
54. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    
55. %Proof
    
56. \newcounter{prf}[section]\setcounter{prf}{0}
    
57. \renewcommand{\theprf}{\arabic{section}.\arabic{prf}}
    
58. \newenvironment{prf}[2][]{%
    
59. \refstepcounter{prf}%
    
60. \ifstrempty{#1}%
    
61. {\mdfsetup{%
    
62. frametitle={%
    
63. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
64. \node[anchor=east,rectangle,fill=red!20]
    
65. {\strut Proof~\theprf};}}
    
66. }%
    
67. {\mdfsetup{%
    
68. frametitle={%
    
69. \tikz[baseline=(current bounding box.east),outer sep=0pt]
    
70. \node[anchor=east,rectangle,fill=red!20]
    
71. {\strut Proof~\thetheo:~#1};}}%
    
72. }%
    
73. \mdfsetup{innertopmargin=10pt,linecolor=red!20,%
    
74. linewidth=2pt,topline=true,%
    
75. frametitleaboveskip=\dimexpr-\ht\strutbox\relax
    
76. }
    
77. \begin{mdframed}[]\relax%
    
78. \label{#2}}{\qed\end{mdframed}}
    
79. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    
80. %Examples
    
81. \begin{document}
    
82. \section{Theorem and lemma examples with title}
    
83. \begin{theo}[Pythagoras' theorem]{thm:pythagoras}
    
84. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the catheti.
    
85. $a^2+b^2=c^2$
    
86. \end{theo}
    
87. In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem (see theorem \ref{thm:pythagoras}), is a relation in Euclidean geometry among the three sides of a right triangle.
    
88. 
    
89. \begin{lem}[B\'ezout's identity]{lem:bezout}
    
90. Let $a$ and $b$ be nonzero integers and let $d$ be their greatest common divisor. Then there exist integers $x$ and $y$ such that:
    
91. $ax+by=d$
    
92. \end{lem}
    
93. This is a reference to Bezout's lemma \ref{lem:bezout}
    
94. 
    
95. 
    
96. \section{Theorem and proof examples without title}
    
97. 
    
98. \begin{theo}{thm:theorem1}
    
99. There exist two irrational numbers $x$, $y$ such that $x^y$ is rational.
    
100. \end{theo}
     
101. 
     
102. \begin{prf}{prf:proof1}
     
103. If $x=y=\sqrt{2}$ is an example, then we are done; otherwise $\sqrt{2}^{\sqrt{2}}$ is irrational, in which case taking $x=\sqrt{2}^{\sqrt{2}}$ and $y=\sqrt{2}$ gives us:
     
104. $\bigg(\sqrt{2}^{\sqrt{2}}\bigg)^{\sqrt{2}}=\sqrt{2}^{\sqrt{2}\sqrt{2}}=\sqrt{2}^{2}=2.$
     
105. \end{prf}
     
106. \end{document}

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