\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{t50\_robbins2 (TMZn5ntDYuWHutfVEKrAEgPgNZjoUMoDMPX)}
\maketitle
Let $ v2\_struct\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_robbins2 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ l2\_robbins1 : {{\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 $ k5\_robbins1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k3\_robbins1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ v4\_lattices : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}((v1\_robbins2~X0){\unicdcdcdhd}(l2\_robbins1~\\
X0))){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(u1\_struct\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(m1\_subset\_1~X2~(u1\_struct\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~\\
(u1\_struct\_0~X0)){\unicdbdjdcd}(k5\_robbins1~X0~(k3\_robbins1~X0~(k5\_robbins1~\\
X0~(k3\_robbins1~X0~X1)~(k3\_robbins1~X0~(k5\_robbins1~X0~(k3\_robbins1~\\
X0~(k3\_robbins1~X0~(k5\_robbins1~X0~X2~X1)))~(k5\_robbins1~X0~X2~X3)))))~\\
(k5\_robbins1~X0~X2~X3)=k5\_robbins1~X0~(k3\_robbins1~X0~(k3\_robbins1~\\
X0~(k5\_robbins1~X0~X2~X1)))~(k5\_robbins1~X0~X2~X3)))))\end{array}
\label{hyp:ehedjdpfchpgcgcgjgogdhcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}((v1\_robbins2~X0){\unicdcdcdhd}(l2\_robbins1~\\
X0))){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(u1\_struct\_0~X0)){\unicdbdjdcd}(k3\_robbins1~\\
X0~(k3\_robbins1~X0~X1)=X1))\end{array}
\label{hyp:ehcdddpfchpgcgcgjgogdhcd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (((\preoh v2\_struct\_0~X0){\unicdcdcdhd}((v4\_lattices~\\
X0){\unicdcdcdhd}(l2\_robbins1~X0))){\unicdcdcdhd}((m1\_subset\_1~X1~(u1\_struct\_0~X0)){\unicdcdcdhd}(\\
m1\_subset\_1~X2~(u1\_struct\_0~X0)))){\unicdbdjdcd}(m1\_subset\_1~(k5\_robbins1~\\
X0~X1~X2)~(u1\_struct\_0~X0))\end{array}
\label{hyp:egehpflgfdpfchpgcgcgjgogdhbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (l2\_robbins1~X0){\unicdbdjdcd}(((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(v1\_robbins2~\\
X0)){\unicdbdjdcd}((\preoh v2\_struct\_0~X0){\unicdcdcdhd}(v4\_lattices~X0)))\end{array}
\label{hyp:dgdgbdpfchpgcgcgjgogdhcd}
\end{equation}
\begin{theorem}\label{thm:ehfdadpfchpgcgcgjgogdhcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v2\_struct\_0~X0){\unicdcdcdhd}((v1\_robbins2~X0){\unicdcdcdhd}(l2\_robbins1~\\
X0))){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~(u1\_struct\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(m1\_subset\_1~X2~(u1\_struct\_0~X0)){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~\\
(u1\_struct\_0~X0)){\unicdbdjdcd}(k5\_robbins1~X0~(k3\_robbins1~X0~(k5\_robbins1~\\
X0~(k3\_robbins1~X0~X1)~(k3\_robbins1~X0~(k5\_robbins1~X0~(k5\_robbins1~\\
X0~X2~X1)~(k5\_robbins1~X0~X2~X3)))))~(k5\_robbins1~X0~X2~X3)=k5\_robbins1~\\
X0~(k3\_robbins1~X0~(k3\_robbins1~X0~(k5\_robbins1~X0~X2~X1)))~(k5\_robbins1~\\
X0~X2~X3)))))\end{array}$$
\end{theorem}
\end{document}
