\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{t56\_setwiseo (TMQxzMPBxVS1otWRpX6VHpTQe2PDovLYzhD)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k5\_finsub\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k11\_setwiseo : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~X0){\unicdbdjdcd}((X1 \in k3\_funct\_2~X0~(k5\_finsub\_1~X0)~(k11\_setwiseo~X0)~X2){\unicdbdjded}\\
(X1=X2)))\end{array}
\label{hyp:ehfdfdpfdhfgehhhjgdhfgpg}
\end{equation}
\begin{theorem}\label{thm:ehfdgdpfdhfgehhhjgdhfgpg}
$$\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~X0){\unicdbdjdcd}\\
({\unicdcdadad} X2 .  (m1\_subset\_1~X2~X0){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~X0){\unicdbdjdcd}\\
(((X1 \in k3\_funct\_2~X0~(k5\_finsub\_1~X0)~(k11\_setwiseo~X0)~X3){\unicdcdcdhd}(X2 \in \\
k3\_funct\_2~X0~(k5\_finsub\_1~X0)~(k11\_setwiseo~X0)~X3)){\unicdbdjdcd}(X1=X2)))))\end{array}$$
\end{theorem}
\end{document}
