\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{t9\_compos\_0 (TMS1FBcusAzw35ZCqx39Huam7i5yktBb3KE)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k5\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (m1\_subset\_1~X0~X1){\unicdbdjdcd}((v1\_xboole\_0~X1){\unicdcdcdid}\\
(X0 \in X1))\end{array}
\label{hyp:ehcdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k5\_xtuple\_0~(k3\_xtuple\_0~X0~\\
X1~X2)=X1\end{array}
\label{hyp:chegfdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k2\_tarski~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3=X0){\unicdcdcdid}(X3=X1)))\end{array}
\label{hyp:egcdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjded}{({{\unicdcdadad} X1 .  {\preoh{{{X1} \in {X0}}}}})}}}\label{hyp:egbdpfihcgpgpgmgfgpfad}
\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:ehjdpfdgpgngahpgdhpfad}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(k2\_tarski~(k3\_xtuple\_0~k6\_numbers~\\
k1\_xboole\_0~k1\_xboole\_0)~(k3\_xtuple\_0~np\_\_1~k1\_xboole\_0~k1\_xboole\_0))){\unicdbdjdcd}\\
(k5\_xtuple\_0~X0=k1\_xboole\_0)\end{array}$$
\end{theorem}
\end{document}
