\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{t86\_funct\_1 (TMHBABLWwPxsQEZJ7Nq61gBWiwZFbmWTRig)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_relat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  ((\\
v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1)){\unicdbdjdcd}(((k9\_xtuple\_0~X0=k9\_xtuple\_0~\\
X1){\unicdcdcdhd}({\unicdcdadad} X2 .  (X2 \in k9\_xtuple\_0~X0){\unicdbdjdcd}(k1\_funct\_1~X0~X2=k1\_funct\_1~\\
X1~X2))){\unicdbdjdcd}(X0=X1)))\end{array}
\label{hyp:ehcdpfggfhogdgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~X2)){\unicdbdjdcd}((X0 \in k9\_xtuple\_0~X1){\unicdbdjdcd}(k1\_funct\_1~\\
(k3\_relat\_1~X1~X2)~X0=k1\_funct\_1~X2~(k1\_funct\_1~X1~X0))))\end{array}
\label{hyp:ehbdddpfggfhogdgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}(v1\_funct\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  (X1=\\
k10\_xtuple\_0~X0){\unicdbdjded}({\unicdcdadad} X2 .  (X2 \in X1){\unicdbdjded}({\unicdcdaddd} X3 .  (X3 \in k9\_xtuple\_0~\\
X0){\unicdcdcdhd}(X2=k1\_funct\_1~X0~X3))))\end{array}
\label{hyp:egddpfggfhogdgehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehidgdpfggfhogdgehpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_relat\_1~X1){\unicdcdcdhd}(v1\_funct\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
((v1\_relat\_1~X2){\unicdcdcdhd}(v1\_funct\_1~X2)){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_relat\_1~\\
X3){\unicdcdcdhd}(v1\_funct\_1~X3)){\unicdbdjdcd}(((X0=k10\_xtuple\_0~X1){\unicdcdcdhd}((k9\_xtuple\_0~\\
X2=X0){\unicdcdcdhd}((k9\_xtuple\_0~X3=X0){\unicdcdcdhd}(k3\_relat\_1~X1~X2=k3\_relat\_1~X1~\\
X3)))){\unicdbdjdcd}(X2=X3))))\end{array}$$
\end{theorem}
\end{document}
