\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{t42\_sublemma (TMbRPWXrxXNJnsx8AZujQfmmrWPbSUMYhHY)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k24\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_cqc\_lang : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k24\_qc\_lang1~X0~(k12\_qc\_lang1~X0)=\\
k1\_xboole\_0)\end{array}
\label{hyp:ehddpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k9\_substut1~X0~(k5\_cqc\_lang~X0)=\\
k1\_xboole\_0)\end{array}
\label{hyp:ehcdpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({{{k5\_cqc\_lang}~{X0}}={{k12\_qc\_lang1}~{X0}}})}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfdgbhdgpfmgbgoghg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0}={X1}})}{\unicdbdjded}{({{({{{r1\_tarski}~{X0}}~{X1}})}{\unicdcdcdhd}{({{{r1\_tarski}~{X1}}~{X0}})}})}}}}\label{hyp:egbdadpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehedcdpfdhfhcgmgfgngngbg}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(r1\_tarski~(k24\_qc\_lang1~X0~(k5\_cqc\_lang~\\
X0))~(k9\_substut1~X0~(k5\_cqc\_lang~X0)))\end{array}$$
\end{theorem}
\end{document}
