\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{l21\_pnproc\_1 (TMbYTfkjhEA4xYRANiEKndchYTADk2YNeEE)}
\maketitle
Let $ k1\_funct\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k1\_pnproc\_1 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k6\_numbers : {\iota}$ be given.
Let $ k16\_pre\_poly : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ c2\_\_arytm\_0 : {\iota}$ be given.
Let $ c1\_\_axioms : {\iota}$ be given.
Let $ k8\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k5\_numbers : {\iota}$ be given.
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k1\_funct\_1}~{({{k16\_pre\_poly}~{X0}})}}~{X1}}={k6\_numbers}}}}\label{hyp:ehfdcdpfahchfgpfahpgmgjh}
\end{equation}
Assume the following.
\begin{equation}
{{c2\_\_arytm\_0}={k6\_numbers}}\label{hyp:egfgpfdgcdpfpfbgchjhehngpfad}
\end{equation}
Assume the following.
\begin{equation}
{{c1\_\_axioms}={k6\_numbers}}\label{hyp:egfgpfdgbdpfpfbgihjgpgngdh}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k1\_pnproc\_1}~{X0}}={{{{k8\_funcop\_1}~{k5\_numbers}}~{X0}}~{k6\_numbers}}}}\label{hyp:egcdpfahogahchpgdgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{{k16\_pre\_poly}~{X0}}={{{{k8\_funcop\_1}~{k5\_numbers}}~{X0}}~{k6\_numbers}}}}\label{hyp:egbdddpfahchfgpfahpgmgjh}
\end{equation}
\begin{theorem}\label{thm:mgcdbdpfahogahchpgdgpfbd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (X0 \in X1){\unicdbdjdcd}(k1\_funct\_1~(k1\_pnproc\_1~X1)~X0=\\
k6\_numbers)\end{array}$$
\end{theorem}
\end{document}
