\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{l9\_ami\_2 (TMXSWLX7RekB4U46Vew4eKDEz6u5t4rShjp)}
\maketitle
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_ami\_2 : {\iota}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_scm\_inst : {\iota}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X0 \in k2\_xboole\_0~X2~(k1\_tarski~\\
X1)){\unicdbdjded}((X0 \in X2){\unicdcdcdid}(X0=X1))\end{array}
\label{hyp:ehbdddgdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_ami\_2}={{{k2\_xboole\_0}~{({{k1\_tarski}~{k5\_numbers}})}}~{k2\_scm\_inst}}}\label{hyp:egbdpfbgngjgpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_xboole\_0}~{X0}}~{X1}}={{{k2\_xboole\_0}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:mgjdpfbgngjgpfcd}
${{k5\_numbers} \in {k1\_ami\_2}}$.
\end{theorem}
\end{document}
