How can I realize this proof tree in LaTeX?
Does anyone have an idea how to achieve this?
[Tex/LaTex] How to create a proof tree with LaTeX
bracketsproof-package
Related Solutions
Another solution, based on listliketab:
\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{listliketab}
\begin{document}
\storestyleof{enumerate}
\begin{listliketab}%
\newcounter{tabenum}\setcounter{tabenum}{0}
\newcommand{\step}{\refstepcounter{tabenum}\thetabenum.}
\noindent \begin{tabular}{L >{$}l <{$}@{\hskip4em}l}
\multicolumn{2}{c}{$ \vdash P\lor\lnot P $} \\
\step & \lnot(P\land\lnot P) & Hyp \\
\step & P & {Hyp}\\
\step & P\lor\lnot P & 2, $\lor$-introduction \\
\step & \bot & 1, 3, contradiction \\
\step & \lnot P & 2, 4, negation-introduction \\
\step & P\lor\lnot P & 5, $\lor$-introduction \\
\step & \bot & 1, 6, contradiction \\
\step & \lnot\lnot(P\lor\lnot P) & 1, 7, negation-introduction \\
\step & P\lor\lnot P & 8, double negation elimination
\end{tabular}
\end{listliketab}
\end{document}
The proof environment is not defined so you get the error
! LaTeX Error: Environment proof undefined.
a common way to define it is to load amsthm when you get a proof heading and qed box at the end.
\documentclass[12pt]{report}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{tikz}
\usepackage{graphicx}
\usepackage{float}
\usepackage{framed}
\usepackage[hang,flushmargin]{footmisc}
\usepackage{amsthm}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[theorem]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}{Definition}[section]
\begin{document}
\begin{proof}
Assume that $R$ decides $\mathrm{HALT}_{\mathrm{TM}}$, and obtain a contradiction. Construct $S$ to decide $A_{\mathrm{TM}}$, where $S$ operates as follows:
$S={}$ ``On input $\langle M, w \rangle$, an encoding of a TM $M$ and a string $w$:
\begin{enumerate}
\item Run TM $r$ on input $\langle M, w \rangle$.
\item If $R$ rejects, reject
\item If $R$ accepts, accept
\item If $M$ has accepted, accept; if $M$ has rejected, reject."
\end{enumerate}
If $R$ decides $\mathrm{HALT}_{\mathrm{TM}}$, then $S$ decides $A_{\mathrm{TM}}$. Because $A_{\mathrm{TM}}$ is undecidable, $\mathrm{HALT}_{\mathrm{TM}}$ must also be undecidable.
\end{proof}
\end{document}
Best Answer