\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=}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t4\_pre\_circ (TMay5msWVkiQWwUCEUTHqBtj2pxAR3XPHcf)}
\maketitle
Let $ k2\_funct\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k7\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_funct\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_pboole : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (k2\_zfmisc\_1~X0~X1=k1\_xboole\_0){\unicdbdjded}((X0=\\
k1\_xboole\_0){\unicdcdcdid}(X1=k1\_xboole\_0))\end{array}
\label{hyp:ehjdadpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (k2\_funct\_5~(k7\_funcop\_1~X0~\\
(k7\_funcop\_1~X1~X2))=k7\_funcop\_1~(k2\_zfmisc\_1~X0~X1)~X2){\unicdcdcdhd}(k4\_funct\_5~\\
(k7\_funcop\_1~X0~(k7\_funcop\_1~X1~X2))=k7\_funcop\_1~(k2\_zfmisc\_1~\\
X1~X0)~X2)\end{array}
\label{hyp:ehgdpfggfhogdgehpfgd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k7\_funcop\_1}~{X0}}~{X1}}={{{k2\_funcop\_1}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfggfhogdgpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_pboole}~{X0}}={{{k7\_funcop\_1}~{X0}}~{k1\_xboole\_0}}}}\label{hyp:egddpfahcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k2\_funcop\_1~X0~X1=k2\_zfmisc\_1~X0~(k1\_tarski~\\
X1)\end{array}
\label{hyp:egcdpfggfhogdgpgahpfbd}
\end{equation}
\begin{theorem}\label{thm:ehedpfahchfgpfdgjgchdg}
${{\unicdcdadad} X0 .  {{{k2\_funct\_5}~{({{{k7\_funcop\_1}~{X0}}~{k1\_xboole\_0}})}}={k1\_xboole\_0}}}$.
\end{theorem}
\end{document}
