\documentclass{article}
\newtheorem{axiom}{Axiom}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
\newcommand\unicdcdcdbg{\cup}
\newcommand\unicdcdbdgd{\setminus}
\newcommand\unicdcdcdjd{\cap}
\newcommand\unicdcdadad{\forall}
\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l31\_cfuncdom (TMcpT9QMEKiJYJQZhxDgsxuRKH5nPkRzpQv)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k2\_xboole\_0~(k2\_tarski~X0~X1)~\\
X2{\unicdcdgdad}k1\_xboole\_0\end{array}
\label{hyp:ehededpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {\preoh{{{v1\_xboole\_0}~{({{{k2\_tarski}~{X0}}~{X1}})}}}}}}\label{hyp:ggdgddpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:mgddbdpfdgggfhogdgegpgng}
$$\begin{array}{c}{\unicdcdaddd} X0 .  (\preoh v1\_xboole\_0~X0){\unicdcdcdhd}({\unicdcdaddd} X1 .  {\unicdcdaddd} X2 .  (X0=k2\_tarski~\\
X1~X2){\unicdcdcdhd}(X1{\unicdcdgdad}X2))\end{array}$$
\end{theorem}
\end{document}
