\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{t85\_mcart\_1 (TMd1a5DMEzFnrdRRBtPCxRV9oGv4dsszNL1)}
\maketitle
Let $ k13\_mcart\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k4\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k14\_mcart\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k15\_mcart\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k16\_mcart\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (k1\_xtuple\_0~(k4\_tarski~X0~X1)=X0){\unicdcdcdhd}(k2\_xtuple\_0~\\
(k4\_tarski~X0~X1)=X1)\end{array}
\label{hyp:ehhdpfngdgbgchehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k4\_tarski~X0~X1=k2\_tarski~(k2\_tarski~X0~\\
X1)~(k1\_tarski~X0)\end{array}
\label{hyp:egfdpfehbgchdhlgjg}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k16\_mcart\_1}~{X0}}={{k2\_xtuple\_0}~{({{k2\_xtuple\_0}~{X0}})}}}}\label{hyp:egbdhdpfngdgbgchehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k15\_mcart\_1}~{X0}}={{k1\_xtuple\_0}~{({{k2\_xtuple\_0}~{X0}})}}}}\label{hyp:egbdgdpfngdgbgchehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k14\_mcart\_1}~{X0}}={{k2\_xtuple\_0}~{({{k1\_xtuple\_0}~{X0}})}}}}\label{hyp:egbdfdpfngdgbgchehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k13\_mcart\_1}~{X0}}={{k1\_xtuple\_0}~{({{k1\_xtuple\_0}~{X0}})}}}}\label{hyp:egbdedpfngdgbgchehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehidfdpfngdgbgchehpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
(k13\_mcart\_1~(k4\_tarski~(k4\_tarski~X0~X1)~X2)=X0){\unicdcdcdhd}((k14\_mcart\_1~\\
(k4\_tarski~(k4\_tarski~X0~X1)~X2)=X1){\unicdcdcdhd}((k15\_mcart\_1~(k4\_tarski~\\
X5~(k4\_tarski~X3~X4))=X3){\unicdcdcdhd}(k16\_mcart\_1~(k4\_tarski~X5~(k4\_tarski~\\
X3~X4))=X4)))\end{array}$$
\end{theorem}
\end{document}
