\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}
\newcommand\uniddcgcg{\lambda}
\newcommand\unicdcdaddd{\exists}
\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{t124\_zfmisc\_1 (TMKsKaJxeBya7uYxJCkgzuyRU5rCcgJ5YCE)}
\maketitle
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_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 .  (X0 \in X1){\unicdbdjdcd}(k3\_xboole\_0~X1~(k1\_tarski~X0)=\\
k1\_tarski~X0)\end{array}
\label{hyp:ehedgdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{r1\_tarski}~{X0}}~{X1}})}{\unicdbdjdcd}{({{{{k3\_xboole\_0}~{X0}}~{X1}}={X0}})}}}}\label{hyp:ehcdidpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k3\_xboole\_0~(k3\_xboole\_0~X0~\\
X1)~X2=k3\_xboole\_0~X0~(k3\_xboole\_0~X1~X2)\end{array}
\label{hyp:ehbdgdpfihcgpgpgmgfgpfbd}
\end{equation}
\begin{theorem}\label{thm:ehbdcdedpfkhggngjgdhdgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((r1\_tarski~X0~X1){\unicdcdcdhd}\\
((k3\_xboole\_0~X1~X2=k1\_tarski~X3){\unicdcdcdhd}(X3 \in X0))){\unicdbdjdcd}(k3\_xboole\_0~X0~\\
X2=k1\_tarski~X3)\end{array}$$
\end{theorem}
\end{document}
