\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{t20\_rlvect\_2 (TMdqTxwC53jDr9PEw71Qr8SMyq57tU7Xu33)}
\maketitle
Let $ v2\_struct\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l2\_algstr\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ k4\_rlvect\_2 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_rlvect\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_rlvect\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k5\_funct\_5 : {\iota}$ be given.
Let $ c2\_\_arytm\_0 : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{\preoh{{({{X0} \in {X1}})}}}{\unicdcdcdhd}{({{v1\_xboole\_0}~{X1}})}}}}\label{hyp:ehhdpfcgpgpgmgfg}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(l2\_algstr\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_rlvect\_2~X1~X0){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(u1\_struct\_0~\\
X0)){\unicdbdjdcd}((k3\_funct\_2~(u1\_struct\_0~X0)~k1\_numbers~X1~X2=k6\_numbers){\unicdbdjded}\\
(\preoh X2 \in k3\_rlvect\_2~X0~X1))))\end{array}
\label{hyp:ehbdjdpfchmgghfgdgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_funct\_5}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfggfhogdgehpffd}
\end{equation}
Assume the following.
\begin{equation}
{{v1\_xboole\_0}~{k1\_xboole\_0}}\label{hyp:ggdgbdpfihcgpgpgmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(l2\_algstr\_0~X0)){\unicdbdjdcd}(m1\_rlvect\_2~\\
(k4\_rlvect\_2~X0)~X0)\end{array}
\label{hyp:egehpflgedpfchmgghfgdgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
{{c2\_\_arytm\_0}={k6\_numbers}}\label{hyp:egfgpfdgcdpfpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(l2\_algstr\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_rlvect\_2~X1~X0){\unicdbdjdcd}((X1=k4\_rlvect\_2~X0){\unicdbdjded}(k3\_rlvect\_2~X0~X1=\\
k1\_xboole\_0)))\end{array}
\label{hyp:egfdpfchmgghfgdgehpfcd}
\end{equation}
\begin{theorem}\label{thm:ehcdadpfchmgghfgdgehpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(l2\_algstr\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~(u1\_struct\_0~X0)){\unicdbdjdcd}(k3\_funct\_2~(u1\_struct\_0~X0)~\\
k1\_numbers~(k4\_rlvect\_2~X0)~X1=k6\_numbers))\end{array}$$
\end{theorem}
\end{document}
