\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}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t77\_cohsp\_1 (TMJSViN7PektztddPsamgH7ka7nKZ4EinPh)}
\maketitle
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ k14\_cohsp\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k14\_cohsp\_1~X0~X1=k2\_xboole\_0~(k2\_zfmisc\_1~\\
X0~(k1\_tarski~np\_\_1))~(k2\_zfmisc\_1~X1~(k1\_tarski~np\_\_2))\end{array}
\label{hyp:ehhdddpfdgpgigdhahpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (k4\_tarski~X0~X1 \in \\
k2\_zfmisc\_1~X2~(k1\_tarski~X3)){\unicdbdjded}((X0 \in X2){\unicdcdcdhd}(X1=X3))\end{array}
\label{hyp:ehbdadgdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k4\_tarski~X0~X1=k2\_tarski~(k2\_tarski~X0~\\
X1)~(k1\_tarski~X0)\end{array}
\label{hyp:egfdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdid}(X3 \in X1)))\end{array}
\label{hyp:egddpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_tarski}~{X0}}~{X1}}={{{k2\_tarski}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:ehhdhdpfdgpgigdhahpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (k4\_tarski~X2~np\_\_2 \in k14\_cohsp\_1~\\
X0~X1){\unicdbdjded}(X2 \in X1)\end{array}$$
\end{theorem}
\end{document}
