\documentclass{article}
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\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{example}{Example}
\newtheorem{conjecture}{Conjecture}
\newcommand\unicdcdgdad{\not=}
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\newcommand\unicdbdjded{\Leftrightarrow}
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\newcommand\preoh[1]{\lnot #1}
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\begin{document}
\title{t13\_binop\_1 (TMEuRpJRqhVkqhB6do65jxLXgRMGjGs3m61)}
\maketitle
Let $ v1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v1\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ v1\_binop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ r6\_binop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}}$ be given.
Let $ k3\_binop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_funct\_1~X1){\unicdcdcdhd}((v1\_funct\_2~X1~(k2\_zfmisc\_1~\\
X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~(k2\_zfmisc\_1~(k2\_zfmisc\_1~\\
X0~X0)~X0))))){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_funct\_1~X2){\unicdcdcdhd}((v1\_funct\_2~X2~(\\
k2\_zfmisc\_1~X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~(k2\_zfmisc\_1~\\
(k2\_zfmisc\_1~X0~X0)~X0))))){\unicdbdjdcd}((r6\_binop\_1~X0~X1~X2){\unicdbdjded}({\unicdcdadad} X3 .  \\
(m1\_subset\_1~X3~X0){\unicdbdjdcd}({\unicdcdadad} X4 .  (m1\_subset\_1~X4~X0){\unicdbdjdcd}({\unicdcdadad} X5 .  \\
(m1\_subset\_1~X5~X0){\unicdbdjdcd}((k3\_binop\_1~X0~X1~X3~(k3\_binop\_1~X0~X2~X4~X5)=\\
k3\_binop\_1~X0~X2~(k3\_binop\_1~X0~X1~X3~X4)~(k3\_binop\_1~X0~X1~X3~X5)){\unicdcdcdhd}\\
(k3\_binop\_1~X0~X1~(k3\_binop\_1~X0~X2~X3~X4)~X5=k3\_binop\_1~X0~X2~(k3\_binop\_1~\\
X0~X1~X3~X5)~(k3\_binop\_1~X0~X1~X4~X5))))))))\end{array}
\label{hyp:ehbdbdpfcgjgogpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  (((v1\_funct\_1~X1){\unicdcdcdhd}\\
((v1\_funct\_2~X1~(k2\_zfmisc\_1~X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~(k2\_zfmisc\_1~X0~X0)~X0))))){\unicdcdcdhd}((m1\_subset\_1~X2~X0){\unicdcdcdhd}\\
(m1\_subset\_1~X3~X0))){\unicdbdjdcd}(m1\_subset\_1~(k3\_binop\_1~X0~X1~X2~X3)~X0)\end{array}
\label{hyp:egehpflgddpfcgjgogpgahpfbd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  ((v1\_funct\_1~X1){\unicdcdcdhd}((v1\_funct\_2~X1~(k2\_zfmisc\_1~\\
X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~(k2\_zfmisc\_1~(k2\_zfmisc\_1~\\
X0~X0)~X0))))){\unicdbdjdcd}((v1\_binop\_1~X1~X0){\unicdbdjded}({\unicdcdadad} X2 .  (m1\_subset\_1~X2~\\
X0){\unicdbdjdcd}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~X0){\unicdbdjdcd}(k3\_binop\_1~X0~X1~X2~X3=k3\_binop\_1~\\
X0~X1~X3~X2))))\end{array}
\label{hyp:egcdpfcgjgogpgahpfbd}
\end{equation}
\begin{theorem}\label{thm:ehbdddpfcgjgogpgahpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (\preoh v1\_xboole\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  ((v1\_funct\_1~X1){\unicdcdcdhd}(\\
(v1\_funct\_2~X1~(k2\_zfmisc\_1~X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X1~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~(k2\_zfmisc\_1~X0~X0)~X0))))){\unicdbdjdcd}({\unicdcdadad} X2 .  ((v1\_funct\_1~\\
X2){\unicdcdcdhd}((v1\_funct\_2~X2~(k2\_zfmisc\_1~X0~X0)~X0){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~(k2\_zfmisc\_1~X0~X0)~X0))))){\unicdbdjdcd}((v1\_binop\_1~X2~X0){\unicdbdjdcd}\\
((r6\_binop\_1~X0~X2~X1){\unicdbdjded}({\unicdcdadad} X3 .  (m1\_subset\_1~X3~X0){\unicdbdjdcd}({\unicdcdadad} X4 .  \\
(m1\_subset\_1~X4~X0){\unicdbdjdcd}({\unicdcdadad} X5 .  (m1\_subset\_1~X5~X0){\unicdbdjdcd}(k3\_binop\_1~\\
X0~X2~(k3\_binop\_1~X0~X1~X3~X4)~X5=k3\_binop\_1~X0~X1~(k3\_binop\_1~X0~X2~\\
X3~X5)~(k3\_binop\_1~X0~X2~X4~X5)))))))))\end{array}$$
\end{theorem}
\end{document}
