\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{l22\_matrixc1 (TMTpZCa3xSxPMKmbGFEccgCGwS6kLni9ecv)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k12\_seq\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_finseqop : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k10\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}}$ be given.
Let $ k29\_binop\_2 : {\iota}$ be given.
Let $ k6\_partfun1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m2\_finseq\_1~X0~k2\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_numbers){\unicdbdjdcd}(k12\_seq\_4~X0~X1=k4\_finseqop~k2\_numbers~k2\_numbers~\\
X0~(k10\_funcop\_1~k2\_numbers~k2\_numbers~k29\_binop\_2~X1~(k6\_partfun1~\\
k2\_numbers))))\end{array}
\label{hyp:mgcdpfdgpgngahmgdhahcd}
\end{equation}
\begin{theorem}\label{thm:mgcdcdpfngbgehchjgihdgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m2\_finseq\_1~\\
X1~k2\_numbers){\unicdbdjdcd}(k12\_seq\_4~X1~X0=k4\_finseqop~k2\_numbers~k2\_numbers~\\
X1~(k10\_funcop\_1~k2\_numbers~k2\_numbers~k29\_binop\_2~X0~(k6\_partfun1~\\
k2\_numbers))))\end{array}$$
\end{theorem}
\end{document}
