\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{l12\_robbins4 (TMYCvfKRNuhfVEGsaTLNCaqsUVD7Hn8DR5c)}
\maketitle
Let $ r1\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k6\_subset\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ np\_\_3 : {\iota}$ be given.
Let $ np\_\_2 : {\iota}$ be given.
Let $ np\_\_1 : {\iota}$ be given.
Let $ r1\_tarski : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{o}}}$ be given.
Let $ k4\_xboole\_0 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  (r1\_tarski~X0~(k4\_xboole\_0~X1~\\
X2)){\unicdbdjdcd}((r1\_tarski~X0~X1){\unicdcdcdhd}(r1\_xboole\_0~X0~X2))\end{array}
\label{hyp:ehbdadgdpfihcgpgpgmgfgpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{\unicdcdadad} X0 .  {{\unicdcdadad} X1 .  {{{{k6\_subset\_1}~{X0}}~{X1}}={{{k4\_xboole\_0}~{X0}}~{X1}}}}}\label{hyp:chfgegfgggjgogjgehjgpgogpflggdpfdhfhcgdhfgehpfbd}
\end{equation}
Assume the following.
\begin{equation}
{{{r1\_tarski}~{({{{k6\_subset\_1}~{np\_\_3}}~{np\_\_2}})}}~{({{{k6\_subset\_1}~{np\_\_3}}~{np\_\_1}})}}\label{hyp:mgfdpfjhfgmgmgpghhbdbd}
\end{equation}
\begin{theorem}\label{thm:mgbdcdpfchpgcgcgjgogdhed}
${{{r1\_xboole\_0}~{({{{k6\_subset\_1}~{np\_\_3}}~{np\_\_2}})}}~{np\_\_1}}$.
\end{theorem}
\end{document}
