\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{t22\_rearran1 (TMSRzXmGryDo5638vEcKuT4vjXLGH1XugC4)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_finset\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_numbers : {\iota}$ be given.
Let $ v1\_rearran1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v2\_rearran1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v3\_rearran1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ m2\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ v1\_partfun1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k5\_card\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_rearran1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ k1\_seq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_rearran1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k20\_rfunct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Let $ r1\_xxreal\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k2\_nat\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_finseq\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k3\_rearran1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ m1\_finseq\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k14\_rfunct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k16\_rfunct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k15\_rfunct\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_rfinseq : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_finset\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}(v1\_finset\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~X0){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_funct\_1~X3){\unicdcdcdhd}(m1\_subset\_1~X3~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~X1~k1\_numbers)))){\unicdbdjdcd}({\unicdcdadad} X4 .  ((v1\_rearran1~X4){\unicdcdcdhd}\\
((v2\_rearran1~X4){\unicdcdcdhd}((v3\_rearran1~X4~(k1\_zfmisc\_1~X0)){\unicdcdcdhd}(m2\_finseq\_1~\\
X4~(k1\_zfmisc\_1~X0))))){\unicdbdjdcd}(((v1\_partfun1~X3~X1){\unicdcdcdhd}(k5\_card\_1~X0=\\
k5\_card\_1~X1)){\unicdbdjdcd}(((X2 \in k1\_funct\_1~X4~np\_\_1){\unicdbdjdcd}(k1\_seq\_1~(k3\_rearran1~\\
X1~X0~X4~X3)~X2=k1\_seq\_1~(k20\_rfunct\_3~X1~X3~X1)~np\_\_1)){\unicdcdcdhd}({\unicdcdadad} X5 .  \\
(m1\_subset\_1~X5~k5\_numbers){\unicdbdjdcd}(((r1\_xxreal\_0~np\_\_1~X5){\unicdcdcdhd}(X2 \in k6\_subset\_1~\\
(k1\_funct\_1~X4~(k2\_nat\_1~X5~np\_\_1))~(k1\_funct\_1~X4~X5))){\unicdbdjdcd}((r1\_xxreal\_0~\\
(k3\_finseq\_1~X4)~X5){\unicdcdcdid}(k1\_seq\_1~(k3\_rearran1~X1~X0~X4~X3)~X2=k1\_seq\_1~\\
(k20\_rfunct\_3~X1~X3~X1)~(k2\_nat\_1~X5~np\_\_1)))))))))))\end{array}
\label{hyp:ehbdfdpfchfgbgchchbgogbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{{m2\_finseq\_1}~{X1}}~{X0}})}{\unicdbdjded}{({{{m1\_finseq\_1}~{X1}}~{X0}})}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpfngcdpfggjgogdhfgbhpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_finset\_1~X0)){\unicdcdcdhd}\\
((v1\_rearran1~X1){\unicdcdcdhd}((v2\_rearran1~X1){\unicdcdcdhd}((v3\_rearran1~X1~(k1\_zfmisc\_1~\\
X0)){\unicdcdcdhd}(m1\_finseq\_1~X1~(k1\_zfmisc\_1~X0)))))){\unicdbdjdcd}((v1\_rearran1~(k2\_rearran1~\\
X0~X1)){\unicdcdcdhd}((v2\_rearran1~(k2\_rearran1~X0~X1)){\unicdcdcdhd}((v3\_rearran1~(k2\_rearran1~\\
X0~X1)~(k1\_zfmisc\_1~X0)){\unicdcdcdhd}(m2\_finseq\_1~(k2\_rearran1~X0~X1)~(k1\_zfmisc\_1~\\
X0)))))\end{array}
\label{hyp:egehpflgcdpfchfgbgchchbgogbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_finset\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}(v1\_finset\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_rearran1~\\
X2){\unicdcdcdhd}((v2\_rearran1~X2){\unicdcdcdhd}((v3\_rearran1~X2~(k1\_zfmisc\_1~X1)){\unicdcdcdhd}(m2\_finseq\_1~\\
X2~(k1\_zfmisc\_1~X1))))){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_funct\_1~X3){\unicdcdcdhd}(m1\_subset\_1~\\
X3~(k1\_zfmisc\_1~(k2\_zfmisc\_1~X0~k1\_numbers)))){\unicdbdjdcd}(k4\_rearran1~\\
X0~X1~X2~X3=k14\_rfunct\_3~X1~(k16\_rfunct\_3~X1~(k15\_rfunct\_3~X1~(k2\_rearran1~\\
X1~X2))~(k3\_rfinseq~(k20\_rfunct\_3~X0~X3~X0)))))))\end{array}
\label{hyp:eggdpfchfgbgchchbgogbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_finset\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}(v1\_finset\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_rearran1~\\
X2){\unicdcdcdhd}((v2\_rearran1~X2){\unicdcdcdhd}((v3\_rearran1~X2~(k1\_zfmisc\_1~X1)){\unicdcdcdhd}(m2\_finseq\_1~\\
X2~(k1\_zfmisc\_1~X1))))){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_funct\_1~X3){\unicdcdcdhd}(m1\_subset\_1~\\
X3~(k1\_zfmisc\_1~(k2\_zfmisc\_1~X0~k1\_numbers)))){\unicdbdjdcd}(k3\_rearran1~\\
X0~X1~X2~X3=k14\_rfunct\_3~X1~(k16\_rfunct\_3~X1~(k15\_rfunct\_3~X1~X2)~\\
(k3\_rfinseq~(k20\_rfunct\_3~X0~X3~X0)))))))\end{array}
\label{hyp:egfdpfchfgbgchchbgogbd}
\end{equation}
\begin{theorem}\label{thm:ehcdcdpfchfgbgchchbgogbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}(v1\_finset\_1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((\preoh v1\_xboole\_0~X1){\unicdcdcdhd}(v1\_finset\_1~X1)){\unicdbdjdcd}({\unicdcdadad} X2 .  (m1\_subset\_1~\\
X2~X0){\unicdbdjdcd}({\unicdcdadad} X3 .  ((v1\_funct\_1~X3){\unicdcdcdhd}(m1\_subset\_1~X3~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~X1~k1\_numbers)))){\unicdbdjdcd}({\unicdcdadad} X4 .  ((v1\_rearran1~X4){\unicdcdcdhd}\\
((v2\_rearran1~X4){\unicdcdcdhd}((v3\_rearran1~X4~(k1\_zfmisc\_1~X0)){\unicdcdcdhd}(m2\_finseq\_1~\\
X4~(k1\_zfmisc\_1~X0))))){\unicdbdjdcd}(((v1\_partfun1~X3~X1){\unicdcdcdhd}(k5\_card\_1~X0=\\
k5\_card\_1~X1)){\unicdbdjdcd}(((X2 \in k1\_funct\_1~(k2\_rearran1~X0~X4)~np\_\_1){\unicdbdjdcd}(\\
k1\_seq\_1~(k4\_rearran1~X1~X0~X4~X3)~X2=k1\_seq\_1~(k20\_rfunct\_3~X1~\\
X3~X1)~np\_\_1)){\unicdcdcdhd}({\unicdcdadad} X5 .  (m1\_subset\_1~X5~k5\_numbers){\unicdbdjdcd}(((r1\_xxreal\_0~\\
np\_\_1~X5){\unicdcdcdhd}(X2 \in k6\_subset\_1~(k1\_funct\_1~(k2\_rearran1~X0~X4)~(k2\_nat\_1~\\
X5~np\_\_1))~(k1\_funct\_1~(k2\_rearran1~X0~X4)~X5))){\unicdbdjdcd}((r1\_xxreal\_0~\\
(k3\_finseq\_1~(k2\_rearran1~X0~X4))~X5){\unicdcdcdid}(k1\_seq\_1~(k4\_rearran1~\\
X1~X0~X4~X3)~X2=k1\_seq\_1~(k20\_rfunct\_3~X1~X3~X1)~(k2\_nat\_1~X5~np\_\_1)))))))))))\end{array}$$
\end{theorem}
\end{document}
