\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{t10\_scmfsa10 (TMGv6W7csnCYn8cMDxaZPDXHq6Uzex6Dbm1)}
\maketitle
Let $ v1\_ami\_2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_scmfsa\_2 : {\iota}$ be given.
Let $ k5\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_scmfsa\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k3\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k10\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_ami\_2~X0){\unicdcdcdhd}(m1\_subset\_1~X0~(u1\_struct\_0~k1\_scmfsa\_2))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  ((v1\_ami\_2~X1){\unicdcdcdhd}(m1\_subset\_1~X1~(u1\_struct\_0~k1\_scmfsa\_2))){\unicdbdjdcd}\\
(k6\_scmfsa\_2~X0~X1=k3\_xtuple\_0~np\_\_1~k1\_xboole\_0~(k10\_finseq\_1~\\
X0~X1)))\end{array}
\label{hyp:ehcdpfdhdgngggdhbgbdad}
\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 .  k3\_xtuple\_0~X0~X1~X2=k4\_tarski~\\
(k4\_tarski~X0~X1)~X2\end{array}
\label{hyp:egedpfihehfhahmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdadpfdhdgngggdhbgbdad}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_ami\_2~X0){\unicdcdcdhd}(m1\_subset\_1~X0~(u1\_struct\_0~k1\_scmfsa\_2))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  ((v1\_ami\_2~X1){\unicdcdcdhd}(m1\_subset\_1~X1~(u1\_struct\_0~k1\_scmfsa\_2))){\unicdbdjdcd}\\
(k5\_xtuple\_0~(k6\_scmfsa\_2~X0~X1)=k1\_xboole\_0))\end{array}$$
\end{theorem}
\end{document}
