\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{t27\_funct\_6 (TMTBL4An3cSx747qHv6aJ2jJyC8E7cyPAPa)}
\maketitle
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k3\_card\_3 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k7\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_funct\_6 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_tarski : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_setfam\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X0{\unicdcdgdad}k1\_xboole\_0){\unicdbdjdcd}(k10\_xtuple\_0~(k2\_funcop\_1~\\
X0~X1)=k1\_tarski~X1)\end{array}
\label{hyp:ehidpfggfhogdgpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k3\_tarski}~{({{k1\_tarski}~{X0}})}}={X0}}}\label{hyp:ehcdfdpfkhggngjgdhdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_setfam\_1}~{({{k1\_tarski}~{X0}})}}={X0}}}\label{hyp:ehbdadpfdhfgehggbgngpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k7\_funcop\_1}~{X0}}~{X1}}={{{k2\_funcop\_1}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflghdpfggfhogdgpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{v1\_relat\_1}~{({{{k2\_zfmisc\_1}~{X0}}~{X1}})}}}}\label{hyp:ggdggdpfchfgmgbgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (v1\_funct\_1~(k7\_funcop\_1~X0~X1)){\unicdcdcdhd}((v1\_funct\_2~\\
(k7\_funcop\_1~X0~X1)~X0~(k1\_tarski~X1)){\unicdcdcdhd}(m1\_subset\_1~(k7\_funcop\_1~\\
X0~X1)~(k1\_zfmisc\_1~(k2\_zfmisc\_1~X0~(k1\_tarski~X1)))))\end{array}
\label{hyp:egehpflghdpfggfhogdgpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(k4\_funct\_6~X0=\\
k1\_setfam\_1~(k10\_xtuple\_0~X0))\end{array}
\label{hyp:egedpfggfhogdgehpfgd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}(k3\_card\_3~X0=\\
k3\_tarski~(k10\_xtuple\_0~X0))\end{array}
\label{hyp:egedpfdgbgchegpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  k2\_funcop\_1~X0~X1=k2\_zfmisc\_1~X0~(k1\_tarski~\\
X1)\end{array}
\label{hyp:egcdpfggfhogdgpgahpfbd}
\end{equation}
\begin{theorem}\label{thm:ehcdhdpfggfhogdgehpfgd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X0{\unicdcdgdad}k1\_xboole\_0){\unicdbdjdcd}((k3\_card\_3~(k7\_funcop\_1~\\
X0~X1)=X1){\unicdcdcdhd}(k4\_funct\_6~(k7\_funcop\_1~X0~X1)=X1))\end{array}$$
\end{theorem}
\end{document}
