\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\_qc\_lang3 (TMckiBjt1ksLWohAEDmyUzsjDxjdbUJKm7Y)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k5\_qc\_lang3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_qc\_lang2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k9\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k13\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(k9\_qc\_lang1~\\
X0)){\unicdbdjdcd}(k5\_qc\_lang3~X0~(k13\_qc\_lang1~X0~X1)=k5\_qc\_lang3~X0~X1))\end{array}
\label{hyp:ehfdfdpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k5\_qc\_lang3~X0~(k12\_qc\_lang1~X0)=\\
k1\_xboole\_0)\end{array}
\label{hyp:ehfdddpfbhdgpfmgbgoghgdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(m1\_subset\_1~(k12\_qc\_lang1~X0)~(\\
k9\_qc\_lang1~X0))\end{array}
\label{hyp:egehpflgbdcdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k1\_qc\_lang2~X0=k13\_qc\_lang1~X0~\\
(k12\_qc\_lang1~X0))\end{array}
\label{hyp:egbdpfbhdgpfmgbgoghgcd}
\end{equation}
\begin{theorem}\label{thm:ehfdgdpfbhdgpfmgbgoghgdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k5\_qc\_lang3~X0~(k1\_qc\_lang2~X0)=\\
k1\_xboole\_0)\end{array}$$
\end{theorem}
\end{document}
