\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{t1\_recdef\_2 (TMSsXoUUo7Z1AWXqG5RkG4LWEsYA86HAoAj)}
\maketitle
Let $ k3\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k4\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k5\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v2\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v2\_xtuple\_0~X0){\unicdbdjdcd}(k3\_xtuple\_0~(k4\_xtuple\_0~X0)~(k5\_xtuple\_0~\\
X0)~(k2\_xtuple\_0~X0)=X0)\end{array}
\label{hyp:cheghdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  v2\_xtuple\_0~(k3\_xtuple\_0~X0~\\
X1~X2)\end{array}
\label{hyp:ggdgcdpfihehfhahmgfgpfad}
\end{equation}
\begin{theorem}\label{thm:ehbdpfchfgdgegfgggpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ({\unicdcdaddd} X1 .  {\unicdcdaddd} X2 .  {\unicdcdaddd} X3 .  X0=k3\_xtuple\_0~\\
X1~X2~X3){\unicdbdjdcd}(X0=k3\_xtuple\_0~(k4\_xtuple\_0~X0)~(k5\_xtuple\_0~X0)~(k2\_xtuple\_0~\\
X0))\end{array}$$
\end{theorem}
\end{document}
