\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{t6\_scm\_inst (TMVaFsCdAQWmf9txfMGePtFbmjty9gQYuzs)}
\maketitle
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_scm\_inst : {\iota}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ k7\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ np\_\_9 : {\iota}$ be given.
Let $ k3\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k12\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k6\_scm\_inst : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k7\_partfun1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k4\_ordinal1 : {\iota}$ be given.
Let $ k5\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_finset\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~X1~X0){\unicdbdjdcd}\\
(k7\_partfun1~X0~(k12\_finseq\_1~X0~X1)~np\_\_1=X1))\end{array}
\label{hyp:ehbdgdpfggjgogdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
{{k5\_numbers}={k4\_ordinal1}}\label{hyp:chfgegfgggjgogjgehjgpgogpflgfdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  k5\_xtuple\_0~(k3\_xtuple\_0~X0~\\
X1~X2)=X1\end{array}
\label{hyp:chegfdpfihehfhahmgfgpfad}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_finset\_1}~{k4\_ordinal1}}}}\label{hyp:ggdghdpfdgbgchegpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(m1\_subset\_1~X1~X0)){\unicdbdjdcd}\\
(m2\_finseq\_1~(k12\_finseq\_1~X0~X1)~X0)\end{array}
\label{hyp:egehpflgbdcdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k3\_scm\_inst){\unicdbdjdcd}(({\unicdcdaddd} X1 .  (m1\_subset\_1~\\
X1~k5\_numbers){\unicdcdcdhd}({\unicdcdaddd} X2 .  (m1\_subset\_1~X2~(k7\_card\_1~np\_\_9)){\unicdcdcdhd}\\
(X0=k3\_xtuple\_0~X2~(k12\_finseq\_1~k5\_numbers~X1)~k1\_xboole\_0))){\unicdbdjdcd}\\
({\unicdcdadad} X1 .  (m1\_subset\_1~X1~k5\_numbers){\unicdbdjdcd}((X1=k6\_scm\_inst~X0){\unicdbdjded}\\
({\unicdcdaddd} X2 .  (m2\_finseq\_1~X2~k5\_numbers){\unicdcdcdhd}((X2=k5\_xtuple\_0~X0){\unicdcdcdhd}\\
(X1=k7\_partfun1~k5\_numbers~X2~np\_\_1))))))\end{array}
\label{hyp:egfdpfdhdgngpfjgogdheh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xboole\_0}~{X0}})}{\unicdbdjdcd}{({{v1\_finset\_1}~{X0}})}}}\label{hyp:dgdgbdpfggjgogdhfgehpfbd}
\end{equation}
\begin{theorem}\label{thm:ehgdpfdhdgngpfjgogdheh}
$$\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k3\_scm\_inst){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k5\_numbers){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~(k7\_card\_1~np\_\_9)){\unicdbdjdcd}\\
((X0=k3\_xtuple\_0~X2~(k12\_finseq\_1~k5\_numbers~X1)~k1\_xboole\_0){\unicdbdjdcd}\\
(k6\_scm\_inst~X0=X1))))\end{array}$$
\end{theorem}
\end{document}
