\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{l3\_rvsum\_2 (TMJQV23aPPZDLHt2CDY1RJWxWaHHz87Qdtr)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k5\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k29\_binop\_2 : {\iota}$ be given.
Let $ k6\_partfun1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_numbers : {\iota}$ be given.
Let $ k5\_binop\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ m1\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k3\_funct\_2 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Let $ k7\_seq\_4 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
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 $ k1\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k2\_zfmisc\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (m1\_subset\_1~X0~k2\_numbers){\unicdbdjdcd}({\unicdcdadad} X1 .  (m1\_subset\_1~\\
X1~k2\_numbers){\unicdbdjdcd}(k3\_funct\_2~k2\_numbers~k2\_numbers~(k7\_seq\_4~X0)~\\
X1=k5\_binop\_2~X0~X1))\end{array}
\label{hyp:ehfdfdpfdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{({{X0} \in {X1}})}{\unicdbdjdcd}{({{{m1\_subset\_1}~{X0}}~{X1}})}}}}\label{hyp:ehbdpfdhfhcgdhfgeh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  ((\preoh v1\_xboole\_0~X0){\unicdcdcdhd}\\
(((v1\_funct\_1~X2){\unicdcdcdhd}((v1\_funct\_2~X2~X0~X1){\unicdcdcdhd}(m1\_subset\_1~X2~(k1\_zfmisc\_1~\\
(k2\_zfmisc\_1~X0~X1))))){\unicdcdcdhd}(m1\_subset\_1~X3~X0))){\unicdbdjdcd}(k3\_funct\_2~X0~\\
X1~X2~X3=k1\_funct\_1~X2~X3)\end{array}
\label{hyp:chfgegfgggjgogjgehjgpgogpflgddpfggfhogdgehpfcd}
\end{equation}
Assume the following.
\begin{equation}
{\preoh{{{v1\_xboole\_0}~{k2\_numbers}}}}\label{hyp:ggdgcdpfogfhngcgfgchdh}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}((v1\_funct\_1~(k7\_seq\_4~X0)){\unicdcdcdhd}((v1\_funct\_2~\\
(k7\_seq\_4~X0)~k2\_numbers~k2\_numbers){\unicdcdcdhd}(m1\_subset\_1~(k7\_seq\_4~\\
X0)~(k1\_zfmisc\_1~(k2\_zfmisc\_1~k2\_numbers~k2\_numbers)))))\end{array}
\label{hyp:egehpflghdpfdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}(k7\_seq\_4~X0=k5\_funcop\_1~k29\_binop\_2~\\
X0~(k6\_partfun1~k2\_numbers))\end{array}
\label{hyp:egedpfdhfgbhpfed}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{({{v1\_xcmplx\_0}~{X0}})}{\unicdbdjded}{({{X0} \in {k2\_numbers}})}}}\label{hyp:egcdpfihdgngahmgihpfad}
\end{equation}
\begin{theorem}\label{thm:mgddpfchghdhfhngpfcd}
$$\begin{array}{c}{\unicdcdadad} X0 .  (v1\_xcmplx\_0~X0){\unicdbdjdcd}({\unicdcdadad} X1 .  (v1\_xcmplx\_0~X1){\unicdbdjdcd}(k1\_funct\_1~\\
(k5\_funcop\_1~k29\_binop\_2~X0~(k6\_partfun1~k2\_numbers))~X1=k5\_binop\_2~\\
X0~X1))\end{array}$$
\end{theorem}
\end{document}
