\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{t2\_ami\_6 (TMJUHNpensws9bL1NPGxgwh8UFJ5vAffdAu)}
\maketitle
Let $ k5\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_compos\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_ami\_3 : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k3\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{{k2\_compos\_1}~{k1\_ami\_3}}={{{{k3\_xtuple\_0}~{k6\_numbers}}~{k1\_xboole\_0}}~{k1\_xboole\_0}}}\label{hyp:ehcdgdpfbgngjgpfdd}
\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}
\begin{theorem}\label{thm:ehcdpfbgngjgpfgd}
${{{k5\_xtuple\_0}~{({{k2\_compos\_1}~{k1\_ami\_3}})}}={k1\_xboole\_0}}$.
\end{theorem}
\end{document}
