\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{t5\_qc\_lang1 (TMPjRHzNjeMQBjMZdUNnbbwTqp4qVWpycLr)}
\maketitle
Let $ m1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k1\_qc\_lang1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (k2\_zfmisc\_1~X0~X1=k1\_xboole\_0){\unicdbdjded}((X0=\\
k1\_xboole\_0){\unicdcdcdid}(X1=k1\_xboole\_0))\end{array}
\label{hyp:ehjdadpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  \preoh (X0{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}((X1{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}\\
(\preoh (k9\_xtuple\_0~(k2\_zfmisc\_1~X0~X1)=X0){\unicdcdcdhd}(k10\_xtuple\_0~(k2\_zfmisc\_1~\\
X0~X1)=X1)))\end{array}
\label{hyp:ehbdgdadpfchfgmgbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{k1\_xboole\_0}}\label{hyp:ggdgbdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{k1\_xboole\_0}={{the}~{({{\uniddcgcg} X0 : {\iota} .  {{v1\_xboole\_0}~{X0}}})}}}\label{hyp:egcdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{m1\_qc\_lang1}~{X0}})}{\unicdbdjdcd}{({{{k1\_qc\_lang1}~{X0}}={{k10\_xtuple\_0}~{X0}}})}}}\label{hyp:egcdpfbhdgpfmgbgoghgbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjded}((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}({\unicdcdaddd} X1 .  \\
(r1\_tarski~k5\_numbers~X1){\unicdcdcdhd}(X0=k2\_zfmisc\_1~k5\_numbers~X1)))\end{array}
\label{hyp:egbdpfbhdgpfmgbgoghgbd}
\end{equation}
\begin{theorem}\label{thm:ehfdpfbhdgpfmgbgoghgbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_qc\_lang1~X0){\unicdbdjdcd}(X0=k2\_zfmisc\_1~k5\_numbers~(k1\_qc\_lang1~\\
X0))\end{array}$$
\end{theorem}
\end{document}
