\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{t47\_funct\_5 (TMW4XhFdBhyLv2Ck9WfXByTMcA7HLenGsMk)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_funct\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_funct\_4 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_funct\_5 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}((r1\_tarski~(k9\_xtuple\_0~X2)~(k2\_zfmisc\_1~X0~X1)){\unicdbdjdcd}(k2\_funct\_4~\\
(k2\_funct\_4~X2)=X2))\end{array}
\label{hyp:ehfdcdpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_relat\_1~X3){\unicdcdcdhd}(v1\_funct\_1~X3)){\unicdbdjdcd}(((r1\_tarski~\\
(k10\_xtuple\_0~X2)~(k1\_funct\_2~X0~X1)){\unicdcdcdhd}((r1\_tarski~(k10\_xtuple\_0~\\
X3)~(k1\_funct\_2~X0~X1)){\unicdcdcdhd}(k2\_funct\_5~X2=k2\_funct\_5~X3))){\unicdbdjdcd}((X0=\\
k1\_xboole\_0){\unicdcdcdid}(X2=X3))))\end{array}
\label{hyp:ehedgdpfggfhogdgehpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}((r1\_tarski~(k10\_xtuple\_0~X2)~(k1\_funct\_2~X0~X1)){\unicdbdjdcd}((k9\_xtuple\_0~\\
(k2\_funct\_5~X2)=k2\_zfmisc\_1~(k9\_xtuple\_0~X2)~X0){\unicdcdcdhd}(k9\_xtuple\_0~\\
(k4\_funct\_5~X2)=k2\_zfmisc\_1~X0~(k9\_xtuple\_0~X2))))\end{array}
\label{hyp:ehcdgdpfggfhogdgehpffd}
\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 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}((v1\_relat\_1~(\\
k2\_funct\_5~X0)){\unicdcdcdhd}(v1\_funct\_1~(k2\_funct\_5~X0)))\end{array}
\label{hyp:egehpflgcdpfggfhogdgehpffd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(k4\_funct\_5~X0=\\
k2\_funct\_4~(k2\_funct\_5~X0))\end{array}
\label{hyp:egedpfggfhogdgehpffd}
\end{equation}
\begin{theorem}\label{thm:ehedhdpfggfhogdgehpffd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  ((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~\\
X2)){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_relat\_1~X3){\unicdcdcdhd}(v1\_funct\_1~X3)){\unicdbdjdcd}(((r1\_tarski~\\
(k10\_xtuple\_0~X2)~(k1\_funct\_2~X0~X1)){\unicdcdcdhd}((r1\_tarski~(k10\_xtuple\_0~\\
X3)~(k1\_funct\_2~X0~X1)){\unicdcdcdhd}(k4\_funct\_5~X2=k4\_funct\_5~X3))){\unicdbdjdcd}((X0=\\
k1\_xboole\_0){\unicdcdcdid}(X2=X3))))\end{array}$$
\end{theorem}
\end{document}
