\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{t67\_finseq\_3 (TMLsuqeMJwDPKLWJtWP72fa8rGWJy9cp7R4)}
\maketitle
Let $ v1\_relat\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k10\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_finseq\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ r1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_xboole\_0 : {{\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){\unicdcdcdhd}(v1\_finseq\_1~X0))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  k10\_xtuple\_0~(k1\_finseq\_3~X0~X1)=k6\_subset\_1~(k10\_xtuple\_0~\\
X0)~X1)\end{array}
\label{hyp:ehgdfdpfggjgogdhfgbhpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (\preoh (\preoh r1\_xboole\_0~X0~X1){\unicdcdcdhd}({\unicdcdadad} X2 .  \preoh (X2 \in \\
X0){\unicdcdcdhd}(X2 \in X1))){\unicdcdcdhd}(\preoh ({\unicdcdaddd} X2 .  (X2 \in X0){\unicdcdcdhd}(X2 \in X1)){\unicdcdcdhd}(r1\_xboole\_0~\\
X0~X1))\end{array}
\label{hyp:ehddpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k6\_subset\_1}~{X0}}~{X1}}={{{k4\_xboole\_0}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (X2=k4\_xboole\_0~X0~X1){\unicdbdjded}({\unicdcdadad} X3 .  \\
(X3 \in X2){\unicdbdjded}((X3 \in X0){\unicdcdcdhd}(\preoh X3 \in X1)))\end{array}
\label{hyp:egfdpfihcgpgpgmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehgdhdpfggjgogdhfgbhpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v1\_relat\_1~X0){\unicdcdcdhd}((v1\_funct\_1~X0){\unicdcdcdhd}(v1\_finseq\_1~X0))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  (k10\_xtuple\_0~(k1\_finseq\_3~X0~X1)=k10\_xtuple\_0~X0){\unicdbdjdcd}\\
(r1\_xboole\_0~X1~(k10\_xtuple\_0~X0)))\end{array}$$
\end{theorem}
\end{document}
