编辑

Question

您可以像添加其他图表一样添加树：直接放在文本中间或作为图形。例如：

\documentclass{article}
\usepackage{forest}

\def\shoe{\ensuremath{\supset}}
\def\nee{\ensuremath{\sim}}
\def\nein{\ensuremath{\neg}}
\def\lub{\ensuremath{\lor}}
  \forestset{
    ass/.append style={
      before computing xy={l=\baselineskip},
      no edge
    },
  }

\begin{document}

  Here is some text referring to figure \ref{fig:tree}.
  \begin{figure}
    \centering
      \begin{forest}
        for tree={
          parent anchor=south,
          child anchor=north,
          align=center,
          math content
        },
    [\nee \exists (D^1 \& C^1), ass, name=n1
        [\nee C^1 f, ass, name=n2
            [\nee D^1 f, ass, name=n3
                [\forall \nee (D^1 \& C^1), ass, name=n4
                    [\nee (D^1 f \& C^1 f), ass, name=n5
                        [\nee D^1 f, name=n6]
                        [\nee C^1 f]
                        ]]]]]                ]
        \foreach \i in {1,...,6}
          \node [xshift=-50pt, anchor=east] at (n\i -| n1) {\i.};
        \foreach \i/\j in {1/{(premise)}, 2/{(premise)}, 3/{(negated conclusion)}, 4/{(1, QE)}, 5/{(4, UQ)}, 6/{(5, \nee \&)}}
          \node [xshift=50pt, anchor=west] at (n\i -| n1) {\j};
      \end{forest}
    \caption{A tree}
    \label{fig:tree}
  \end{figure}

\end{document}

请注意，我必须添加math content才能编译，因此输出并不理想，但它演示了如何将其放入普通的article。如果您不想让树木移动（浮动），请不要使用环境figure并将它们放在文本中所需的适当位置。

树如图

或者你可以说

\documentclass{article}
\usepackage{forest,standalone}

\def\shoe{\ensuremath{\supset}}
\def\nee{\ensuremath{\sim}}
\def\nein{\ensuremath{\neg}}
\def\lub{\ensuremath{\lor}}
  \forestset{
    ass/.append style={
      before computing xy={l=\baselineskip},
      no edge
    },
  }

\begin{document}

Some text and now a tree:
\input{tree1}% if tree1.tex is the first tree
\begin{figure}
  \input{tree2}
  \caption{Tree 2}\label{fig:tree2}
\end{figure}
Some more text.

\end{document}

然后您不必触碰现有的树文件或乱七八糟地复制粘贴。

编辑

这是使用第二种方法的正确示例。（请注意，一棵树在环境中figure，而另一棵树不在。这只是为了说明：我不推荐这样做。如果你确实混合了它们，你应该使用capt-of或caption标题来一致地标记非浮动树。）

假设tree1.tex包含：

  \documentclass[tikz,border=5pt,multi]{standalone}
  \usepackage{prooftrees}
  \standaloneenv{forest}
  \newcommand*{\shoe}{\ensuremath{\supset}}
  \newcommand*{\nee}{\ensuremath{\mathord{\sim}}}
  \newcommand*{\nein}{\ensuremath{\mathord{\neg}}}
  \newcommand*{\lub}{\ensuremath{\lor}}
  \newcommand*{\lamp}{\ensuremath{\mathbin{\&}}}
  \begin{document}
    \begin{forest}
      proof tree,
      for tree={math content},
      [, phantom
        [\nee \exists (D^1 \lamp C^1), just=(premise)
          [\nee C^1 f, just=(premise)
            [\nee D^1 f, just=(negated conclusion)
              [\forall \nee (D^1 \lamp C^1), just={(1, QE)}
                [\nee (D^1 f \lamp C^1 f), just={(4, UQ)}
                  [\nee D^1 f, just={(5, \nee \lamp)}
                  ]
                  [\nee C^1 f
                  ]
                ]
              ]
            ]
          ]
        ]
      ]
    \end{forest}
  \end{document}

并tree2.tex包含：

  \documentclass[tikz,border=5pt,multi]{standalone}
  \usepackage{prooftrees}
  \usepackage{amssymb}
  \usepackage{mathtools,turnstile}
  \newcommand*{\tnot}{\ensuremath{\mathord{\sim}}}
  \begin{document}
    \begin{forest}
      proof tree,
      [{$(\exists x)(\forall y)(Py \equiv x = y) \vdash (\exists x)((\forall y)(Py \supset x = y) \,\&\, Px)$}
        [{$(\exists x)(\forall y)(Py \equiv x = y)\ \checkmark a$}, just={Premise}
          [{$\tnot (\exists x)((\forall y)(Py \supset x = y) \,\&\, Px)\ \backslash a$}, just={Conclusion negated}
            [{$(\forall y)(Py \equiv a = y)\ \backslash a,b$}, just={From 1}
              [{$\tnot ((\forall y)(Py \supset a = y) \,\&\, Pa)\ \checkmark$}, just={From 2}
                [{$Pa \equiv a = a\ \checkmark$}, just={From 3}
                  [{$Pa$}, just={From 5}
                    [{$a = a$}, just={From 5}
                      [{$\tnot (\forall y)(Py \supset a = y)\ \checkmark b$}, just={From 4}
                        [{$\tnot (Pb \supset a = b)\ \checkmark$}, just={From 8}
                          [{$Pb$}, just={From 9}
                            [{$a \neq b$}, just={From 9}
                              [{$Pb \equiv a = b\ \checkmark$}, just={From 3}
                                [{$Pb$}, just={From 12}
                                  [{$a = b$}, just={From 12}
                                    [{$a \neq a$}, just={From 11,14}, close
                                    ]
                                  ]
                                ]
                                [{$\tnot Pb$}
                                  [{$a \neq b$}, close
                                  ]
                                ]
                              ]
                            ]
                          ]
                        ]
                      ]
                      [{$\tnot Pa$}, close
                      ]
                    ]
                  ]
                  [{$\tnot Pa$}
                    [{$a \neq a$}, close
                    ]
                  ]
                ]
              ]
            ]
          ]
        ]
      ]
    \end{forest}
  \end{document}

然后我们可以写：

\documentclass{article}
\usepackage{prooftrees,standalone,kantlipsum}
\newcommand*{\shoe}{\ensuremath{\supset}}
\newcommand*{\nee}{\ensuremath{\mathord{\sim}}}
\newcommand*{\nein}{\ensuremath{\mathord{\neg}}}
\newcommand*{\lub}{\ensuremath{\lor}}
\newcommand*{\lamp}{\ensuremath{\mathbin{\&}}}
\usepackage{amssymb}
\usepackage{mathtools,turnstile}
\newcommand*{\tnot}{\ensuremath{\mathord{\sim}}}
\begin{document}

\title{Title}
\author{Me}
\maketitle

\kant[1]

This proof is best illustrated by the following tree which is simply typeset:

\begin{center}
  \emph{here}\par
  \input{tree1}
\end{center}

\kant[2-3]

It is time for another tree.
See figure \ref{fig:tree2} (which is \emph{there} and \emph{not} here) for details of this problem and its solution.
\begin{figure}
  \input{tree2}
  \caption{A tree}
  \label{fig:tree2}
\end{figure}

\kant[3]

\end{document}

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