\documentclass{article}
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\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{t109\_zfmisc\_1 (TMWSTy9zDLv3kFttcyPin8j7LNbFE78Kzpu)}
\maketitle
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ 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\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (k2\_zfmisc\_1~(k2\_xboole\_0~X0~\\
X1)~X2=k2\_xboole\_0~(k2\_zfmisc\_1~X0~X2)~(k2\_zfmisc\_1~X1~X2)){\unicdcdcdhd}(k2\_zfmisc\_1~\\
X2~(k2\_xboole\_0~X0~X1)=k2\_xboole\_0~(k2\_zfmisc\_1~X2~X0)~(k2\_zfmisc\_1~\\
X2~X1))\end{array}
\label{hyp:ehjdhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k2\_tarski~X0~X1=k2\_xboole\_0~(k1\_tarski~\\
X0)~(k1\_tarski~X1)\end{array}
\label{hyp:ehbdpffgogfhngdhfgehbd}
\end{equation}
\begin{theorem}\label{thm:ehbdadjdpfkhggngjgdhdgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (k2\_zfmisc\_1~(k2\_tarski~X0~X1)~\\
X2=k2\_xboole\_0~(k2\_zfmisc\_1~(k1\_tarski~X0)~X2)~(k2\_zfmisc\_1~(\\
k1\_tarski~X1)~X2)){\unicdcdcdhd}(k2\_zfmisc\_1~X2~(k2\_tarski~X0~X1)=k2\_xboole\_0~\\
(k2\_zfmisc\_1~X2~(k1\_tarski~X0))~(k2\_zfmisc\_1~X2~(k1\_tarski~X1)))\end{array}$$
\end{theorem}
\end{document}
