\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}
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\newcommand\unicdcdadjd{\notin}
\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\unicdcdidid{\not\subseteq}
\newcommand\preoh[1]{\lnot #1}
\newcommand\mname[1]{{\mathsf{#1}}}
\begin{document}
\title{l7\_qc\_lang1 (TMTqZkmSsJBkbKatBcwfwcxZ2wftBbSEnK3)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_4 : {\iota}$ be given.
Let $ k1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_5 : {\iota}$ be given.
Let $ np\_\_6 : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((r1\_tarski~X0~X1){\unicdcdcdhd}\\
(r1\_tarski~X2~X3)){\unicdbdjdcd}(r1\_tarski~(k2\_zfmisc\_1~X0~X2)~(k2\_zfmisc\_1~\\
X1~X3))\end{array}
\label{hyp:ehjdgdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{r1\_tarski}~{({{k1\_tarski}~{X0}})}}~{({{{k2\_tarski}~{X0}}~{X1}})}}}}\label{hyp:ehhdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (r1\_tarski~X0~X1){\unicdbdjdcd}(r1\_tarski~\\
X0~(k2\_xboole\_0~X2~X1))\end{array}
\label{hyp:ehbdadpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{r1\_tarski}~{X0}}~{X0}}}}\label{hyp:chfgggmgfgihjgghjgehjhpfchbdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(k2\_qc\_lang1~X0=k2\_xboole\_0~(k2\_zfmisc\_1~\\
(k1\_tarski~np\_\_6)~k5\_numbers)~(k2\_zfmisc\_1~(k2\_tarski~np\_\_4~np\_\_5)~\\
(k1\_qc\_lang1~X0)))\end{array}
\label{hyp:egddpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_xboole\_0}~{X0}}~{X1}}={{{k2\_xboole\_0}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k2\_tarski}~{X0}}~{X1}}={{{k2\_tarski}~{X1}}~{X0}}}}}\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgcdpfehbgchdhlgjg}
\end{equation}
\begin{theorem}\label{thm:mghdpfbhdgpfmgbgoghgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}((r1\_tarski~(k2\_zfmisc\_1~(k1\_tarski~\\
np\_\_4)~(k1\_qc\_lang1~X0))~(k2\_qc\_lang1~X0)){\unicdcdcdhd}((r1\_tarski~(k2\_zfmisc\_1~\\
(k1\_tarski~np\_\_5)~(k1\_qc\_lang1~X0))~(k2\_qc\_lang1~X0)){\unicdcdcdhd}(r1\_tarski~\\
(k2\_zfmisc\_1~(k1\_tarski~np\_\_6)~k5\_numbers)~(k2\_qc\_lang1~X0))))\end{array}$$
\end{theorem}
\end{document}
