\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{l46\_algstr\_1 (TMLPYJFPRGDuedvcXNRux1SMmC8yCDik79a)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ u1\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_algstr\_1 : {\iota}$ be given.
Let $ k6\_algstr\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k4\_struct\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ l5\_algstr\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l4\_algstr\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l4\_struct\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l2\_struct\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l3\_struct\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v29\_algstr\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
{{k6\_numbers}={k1\_xboole\_0}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}(k6\_algstr\_0~\\
k3\_algstr\_1~X0~(k4\_struct\_0~k3\_algstr\_1)=k4\_struct\_0~k3\_algstr\_1)\end{array}
\label{hyp:mgedfdpfbgmghgdhehchpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}((k6\_algstr\_0~k3\_algstr\_1~X0~X1=\\
k6\_algstr\_0~k3\_algstr\_1~X0~X2){\unicdbdjdcd}((X0=k4\_struct\_0~k3\_algstr\_1){\unicdcdcdid}\\
(X1=X2)))))\end{array}
\label{hyp:mgedddpfbgmghgdhehchpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
(m1\_subset\_1~X1~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}(\preoh (X0{\unicdcdgdad}k4\_struct\_0~\\
k3\_algstr\_1){\unicdcdcdhd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}\\
(k6\_algstr\_0~k3\_algstr\_1~X2~X0{\unicdcdgdad}X1))))\end{array}
\label{hyp:mgedcdpfbgmghgdhehchpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{k6\_numbers}={{k4\_struct\_0}~{k3\_algstr\_1}}}\label{hyp:mgddhdpfbgmghgdhehchpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{l5\_algstr\_0}~{X0}})}{\unicdbdjdcd}{({{({{l4\_algstr\_0}~{X0}})}{\unicdcdcdhd}{({{l4\_struct\_0}~{X0}})}})}}}\label{hyp:egehpfmgfdpfbgmghgdhehchpfad}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{l4\_struct\_0}~{X0}})}{\unicdbdjdcd}{({{({{l2\_struct\_0}~{X0}})}{\unicdcdcdhd}{({{l3\_struct\_0}~{X0}})}})}}}\label{hyp:egehpfmgedpfdhehchfhdgehpfad}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (l2\_struct\_0~X0){\unicdbdjdcd}(m1\_subset\_1~(k4\_struct\_0~X0)~(u1\_struct\_0~\\
X0))\end{array}
\label{hyp:egehpflgedpfdhehchfhdgehpfad}
\end{equation}
Assume the following.
\begin{equation}
{{({{v29\_algstr\_0}~{k3\_algstr\_1}})}{\unicdcdcdhd}{({{l5\_algstr\_0}~{k3\_algstr\_1}})}}\label{hyp:egehpflgddpfbgmghgdhehchpfbd}
\end{equation}
\begin{theorem}\label{thm:mgedgdpfbgmghgdhehchpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~(u1\_struct\_0~k3\_algstr\_1)){\unicdbdjdcd}(k6\_algstr\_0~\\
k3\_algstr\_1~(k4\_struct\_0~k3\_algstr\_1)~X0=k4\_struct\_0~k3\_algstr\_1)\end{array}$$
\end{theorem}
\end{document}
