\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{t21\_borsuk\_7 (TMbcDf2Tt5BhyEtUT5629hjNNoa8ZSgdLab)}
\maketitle
Let $ v1\_xcmplx\_0 : {{\iota}{\unicdbdjdcd}{o}}$ be given.
Let $ k9\_xtuple\_0 : {{\iota}{\unicdbdjdcd}{\iota}}$ be given.
Let $ k1\_borsuk\_7 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}}}$ be given.
Let $ k1\_enumset1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}$ be given.
Let $ k1\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k16\_funcop\_1 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}$ be given.
Let $ k4\_funct\_4 : {{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{{\iota}{\unicdbdjdcd}{\iota}}}}}$ be given.
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
k9\_xtuple\_0~(k1\_funct\_4~(k1\_funct\_4~(k16\_funcop\_1~X0~X3)~(k16\_funcop\_1~\\
X1~X4))~(k16\_funcop\_1~X2~X5))=k1\_enumset1~X0~X1~X2\end{array}
\label{hyp:ehbdbdpfcgghggfhogdgbded}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  k4\_funct\_4~X0~X1~X2~\\
X3=k1\_funct\_4~(k16\_funcop\_1~X0~X2)~(k16\_funcop\_1~X1~X3)\end{array}
\label{hyp:egedpfggfhogdgehpfed}
\end{equation}
Assume the following.
\begin{equation}
\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
k1\_borsuk\_7~X0~X1~X2~X3~X4~X5=k1\_funct\_4~(k4\_funct\_4~X0~X1~X3~X4)~(\\
k16\_funcop\_1~X2~X5)\end{array}
\label{hyp:egbdpfcgpgchdhfhlgpfhd}
\end{equation}
\begin{theorem}\label{thm:ehcdbdpfcgpgchdhfhlgpfhd}
$$\begin{array}{c}{\unicdcdadad} X0 .  {\unicdcdadad} X1 .  {\unicdcdadad} X2 .  {\unicdcdadad} X3 .  {\unicdcdadad} X4 .  {\unicdcdadad} X5 .  \\
(v1\_xcmplx\_0~X5){\unicdbdjdcd}(k9\_xtuple\_0~(k1\_borsuk\_7~X0~X1~X5~X2~X3~X4)=k1\_enumset1~\\
X0~X1~X5)\end{array}$$
\end{theorem}
\end{document}
