\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}
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\newcommand\unicdbdjdcd{\Rightarrow}
\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
\newcommand\unicdcdcdid{\lor}
\newcommand\unicdcdcdhd{\land}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
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\begin{document}
\title{t31\_substut1 (TMUfSZmUA75XLdy4knqMnUNQBpSvWBaEPWF)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k16\_substut1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k19\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k37\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k21\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k14\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ v6\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k28\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k29\_substut1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k16\_substut1~\\
X0)){\unicdbdjdcd}((v6\_substut1~X1~X0){\unicdbdjdcd}(k37\_substut1~X0~X1=k14\_qc\_lang1~X0~\\
(k37\_substut1~X0~(k28\_substut1~X0~X1))~(k37\_substut1~X0~(k29\_substut1~\\
X0~X1)))))\end{array}
\label{hyp:ehddadpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k16\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k16\_substut1~X0)){\unicdbdjdcd}((k19\_substut1~\\
X0~X1=k19\_substut1~X0~X2){\unicdbdjdcd}(k29\_substut1~X0~(k21\_substut1~X0~X1~\\
X2)=X2))))\end{array}
\label{hyp:ehbdjdpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k16\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k16\_substut1~X0)){\unicdbdjdcd}((k19\_substut1~\\
X0~X1=k19\_substut1~X0~X2){\unicdbdjdcd}(k28\_substut1~X0~(k21\_substut1~X0~X1~\\
X2)=X1))))\end{array}
\label{hyp:ehbdidpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k16\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k16\_substut1~X0)){\unicdbdjdcd}((k19\_substut1~\\
X0~X1=k19\_substut1~X0~X2){\unicdbdjdcd}(v6\_substut1~(k21\_substut1~X0~X1~X2)~\\
X0))))\end{array}
\label{hyp:ehbdddpfdhfhcgdhehfhehbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((m1\_qc\_lang1~X0){\unicdcdcdhd}((m1\_subset\_1~\\
X1~(k16\_substut1~X0)){\unicdcdcdhd}(m1\_subset\_1~X2~(k16\_substut1~X0)))){\unicdbdjdcd}(\\
m1\_subset\_1~(k21\_substut1~X0~X1~X2)~(k16\_substut1~X0))\end{array}
\label{hyp:egehpflgcdbdpfdhfhcgdhehfhehbd}
\end{equation}
\begin{theorem}\label{thm:ehddbdpfdhfhcgdhehfhehbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k16\_substut1~\\
X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k16\_substut1~X0)){\unicdbdjdcd}((k19\_substut1~\\
X0~X1=k19\_substut1~X0~X2){\unicdbdjdcd}(k37\_substut1~X0~(k21\_substut1~X0~X1~\\
X2)=k14\_qc\_lang1~X0~(k37\_substut1~X0~X1)~(k37\_substut1~X0~X2)))))\end{array}$$
\end{theorem}
\end{document}
