\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}
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\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t8\_scmyciel (TMQmqRvmt6qyMafYRuKDvVhWLxEHm9sUmkH)}
\maketitle
Let $ k3\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k3\_tarski}~{({{k1\_zfmisc\_1}~{X0}})}}={X0}}}\label{hyp:ehidbdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  k1\_zfmisc\_1~(k1\_tarski~X0)=k2\_tarski~k1\_xboole\_0~\\
(k1\_tarski~X0)\end{array}
\label{hyp:ehcdedpfkhggngjgdhdgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehidpfdhdgngjhdgjgfgmg}
$$\begin{array}{c}{\unicdcdadad} X0 .  k3\_tarski~(k2\_tarski~k1\_xboole\_0~(k1\_tarski~X0))=\\
k1\_tarski~X0\end{array}$$
\end{theorem}
\end{document}
