\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}
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\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{t110\_member\_1 (TMSTepqr6eFc7TMRnw8daycaq1KdudZPntf)}
\maketitle
Let $ v2\_membered : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k14\_member\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k12\_member\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_member\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v2\_membered~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v2\_membered~X1){\unicdbdjdcd}(k6\_member\_1~\\
(k12\_member\_1~X0~X1)=k12\_member\_1~(k6\_member\_1~X0)~(k6\_member\_1~\\
X1)))\end{array}
\label{hyp:ehidfdpfngfgngcgfgchpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v2\_membered~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v2\_membered~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v2\_membered~X2){\unicdbdjdcd}(k14\_member\_1~(k12\_member\_1~X0~X1)~X2=k12\_member\_1~\\
X0~(k14\_member\_1~X1~X2))))\end{array}
\label{hyp:ehbdadidpfngfgngcgfgchpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v2\_membered~X0){\unicdcdcdhd}(v2\_membered~X1)){\unicdbdjdcd}(\\
v2\_membered~(k12\_member\_1~X0~X1))\end{array}
\label{hyp:ggdgedbdpfngfgngcgfgchpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v2\_membered}~{X0}})}{\unicdbdjdcd}{({{v2\_membered}~{({{k6\_member\_1}~{X0}})}})}}}\label{hyp:egehpflggdpfngfgngcgfgchpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v2\_membered~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v2\_membered~X1){\unicdbdjdcd}(k14\_member\_1~\\
X0~X1=k12\_member\_1~X0~(k6\_member\_1~X1)))\end{array}
\label{hyp:egbdbdpfngfgngcgfgchpfbd}
\end{equation}
\begin{theorem}\label{thm:ehbdbdadpfngfgngcgfgchpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v2\_membered~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v2\_membered~X1){\unicdbdjdcd}({\unicdcdadad} X2 .  \\
(v2\_membered~X2){\unicdbdjdcd}(k14\_member\_1~(k14\_member\_1~X0~X1)~X2=k14\_member\_1~\\
X0~(k12\_member\_1~X1~X2))))\end{array}$$
\end{theorem}
\end{document}
