# Tree with some nodes and leafs in arbitrary places [closed]

I need to draw a quite abstract tree. This tree will visualize proof substituting in natural deduction (proof theory). I have a deduction (let's call it 'D1') of A with open assumption B and a deduction (let's call it 'D2') of B with open assumption C. Each deduction can be visualized with a tree with its leafs labeled by the proof's assumption and its root labeled by the proof's conclusion.

So, proof substitution can be visualized with substituting all leafs B of tree D1 with the tree of D2. This is like adding D2 as a subtree with root B and leafs C on each leaf B of D1.

What I need to do is draw the tree of the proof that results after the substitution.

I tried this:

\documentclass[a4paper]{scrartcl}
\usepackage{fancybox}
\usepackage{tikz}

\title{MergeSort-RecursionTree}
\author{Manuel Kirsch}
\date{}
\begin{document}

\begin{tikzpicture}[level/.style={sibling distance=5cm/#1}]
\node (z){$A$}
child {node (a) {$\vdots$}
child {node (b) {$B$}
child {node (A) {$\vdots$}
child {node (B) {$C$}}
child {node (G) {$C$}}
}
child {node (J) {$\vdots$}
child {node (K) {$C$}}
child {node (L) {$C$}}
}
}
child {node (g) {$B$}
child {node (A) {$\vdots$}
child {node (B) {$C$}}
child {node (G) {$C$}}
}
child {node (J) {$\vdots$}
child {node (K) {$C$}}
child {node (L) {$C$}}
}
}
}
child {node (j) {$\vdots$}
child {node (k) {$B$}
child {node (A) {$\vdots$}
child {node (B) {$C$}}
child {node (G) {$C$}}
}
child {node (J) {$\vdots$}
child {node (K) {$C$}}
child {node (L) {$C$}}
}}
child {node (l) {$B$}
child {node (A) {$\vdots$}
child {node (B) {$C$}}
child {node (G) {$C$}}
}
child {node (J) {$\vdots$}
child {node (K) {$C$}}
child {node (L) {$C$}}
}}
};
\path (a) -- (j) node [midway] {$\cdots$};
\path (b) -- (g) node [midway] {$\cdots$};
\path (k) -- (l) node [midway] {$\cdots$};
\path (k) -- (g) node [midway] {$\cdots$};
\path (A) -- (J) node [midway] {$\cdots$};
\path (K) -- (G) node [midway] {$\cdots$};
\path (B) -- (G) node [midway] {$\cdots$};
\path (K) -- (L) node [midway] {$\cdots$};
\end{tikzpicture}

\end{document}


But except that some lower edges cross, this kind of tree gives to the reader the impression that all nodes B are in the same level and the same for C, too. In addition, it is not clear that the tree is not binary.

Is there a more abstract tree such that it visualizes only the necessary information and does not add any further impressions about its structure to the reader? What I want to emphasize about this tree is that all nodes B have the same subtree.

To be honest, I can't even imagine such a tree. I don't know how to visualize arbitrary number of children for each node or arbitrary lengths for the paths from the root to each leaf. So, any ideas/suggestions on that would be really helpful! Code for such a tree would be highly appreciated, since don't have much experience with tikz.

## closed as off-topic by Adam Liter, jub0bs, Thorsten, Jesse, Peter JanssonMar 14 '14 at 15:52

• This question does not fall within the scope of TeX, LaTeX or related typesetting systems as defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

• Can you please provide a hand drawn image of the general layout you desire for the tree? – Gonzalo Medina Mar 2 '14 at 0:44
• @GonzaloMedina I would, but the problem is I cannot imagine such a tree. I don't know how to visualize arbitrary number of children for each node or arbitrary lengths for each path from the root to each leaf. Part of this question is also how to picture such an abstraction. Edited my post. – frabala Mar 2 '14 at 0:53
• If the question is about how one would even draw the abstraction to begin with, it's honestly probably off topic here. That being said, @cfr might have an idea. Maybe they will see this. – Adam Liter Mar 14 '14 at 5:56
• This question appears to be off-topic because it is about how to, in principle, draw a tree illustrating an abstract deduction and not about how to draw a particular tree in (La)TeX. – Adam Liter Mar 14 '14 at 14:44