\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{t25\_arytm\_3 (TMRXbMJQBhmhgGMpXGHjSpouxewAqaJeHyQ)}
\maketitle
Let $ v3\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ v7\_ordinal1 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k1\_xboole\_0 : {\iota}$ be given.
Let $ k4\_arytm\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k3\_arytm\_3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k6\_ordinal3 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v3\_ordinal1~X1){\unicdcdcdhd}(v7\_ordinal1~X1)){\unicdbdjdcd}((X0{\unicdcdgdad}k1\_xboole\_0){\unicdbdjdcd}((k3\_arytm\_3~\\
X1~X0{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}(k6\_ordinal3~X0~(k3\_arytm\_3~X1~X0){\unicdcdgdad}k1\_xboole\_0))))\end{array}
\label{hyp:ehbdidpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v3\_ordinal1~X1){\unicdcdcdhd}(v7\_ordinal1~X1)){\unicdbdjdcd}(k4\_arytm\_3~X0~X1=k6\_ordinal3~\\
X0~(k3\_arytm\_3~X0~X1)))\end{array}
\label{hyp:eggdpfbgchjhehngpfdd}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  (((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdcdcdhd}\\
((v3\_ordinal1~X1){\unicdcdcdhd}(v7\_ordinal1~X1))){\unicdbdjdcd}(k3\_arytm\_3~X0~X1=k3\_arytm\_3~\\
X1~X0)\end{array}
\label{hyp:dgpgngngfhehbgehjgghjgehjhpflgddpfbgchjhehngpfdd}
\end{equation}
\begin{theorem}\label{thm:ehcdfdpfbgchjhehngpfdd}
$$\begin{array}{c}{\unicdcdadad} X0 .  ((v3\_ordinal1~X0){\unicdcdcdhd}(v7\_ordinal1~X0)){\unicdbdjdcd}({\unicdcdadad} X1 .  \\
((v3\_ordinal1~X1){\unicdcdcdhd}(v7\_ordinal1~X1)){\unicdbdjdcd}(\preoh (X0{\unicdcdgdad}k1\_xboole\_0){\unicdcdcdhd}(k4\_arytm\_3~\\
X0~X1=k1\_xboole\_0)))\end{array}$$
\end{theorem}
\end{document}
