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Assume the theorem is false. We know the following: 
\begin{equation} \label{4}
(\forall v_{0})\ \operatorname{sv_{1}}\left(v_{0}\right) = \operatorname{sv_{2}}\left(v_{0}\right) \implies \operatorname{SINGLEVALUED}\left(v_{0}\right)
\end{equation}
\begin{equation} \label{6}
(\forall v_{0},v_{1})\ \left\langle v_{0},v_{1}\right\rangle  \in \text{\emph{POWER}} \implies \mathscr{P}(v_{0}) = v_{1}
\end{equation}We then substitute $v_{64}$ for $v_{0}$ to obtain 
\begin{equation} \label{13}
(\forall v_{1},v_{64})\ \left\langle v_{64},v_{1}\right\rangle  \in \text{\emph{POWER}} \implies \mathscr{P}(v_{64}) = v_{1}
\end{equation}In $(\forall v_{0})\ \operatorname{U}\left(\mathscr{P}(v_{0})\right) = v_{0}$ we substitute $v_{64}$ for $v_{0}$ to obtain 
\begin{equation*} \label{14}
(\forall v_{64})\ \operatorname{U}\left(\mathscr{P}(v_{64})\right) = v_{64}
\end{equation*}which together with (\ref{13}) yields 
\begin{equation*} \label{15}
(\forall v_{1},v_{64})\ \left\langle v_{64},v_{1}\right\rangle  \in \text{\emph{POWER}} \implies \operatorname{U}\left(v_{1}\right) = v_{64}
\end{equation*}We then substitute $v_{0}$ for $v_{64}$ to obtain 
\begin{equation} \label{16}
(\forall v_{0},v_{1})\ \left\langle v_{0},v_{1}\right\rangle  \in \text{\emph{POWER}} \implies \operatorname{U}\left(v_{1}\right) = v_{0}
\end{equation}We know the following: 
\begin{equation*} \label{1}
(\forall v_{0})\ v_{0} \subseteq V \times V \;\wedge\; \operatorname{SINGLEVALUED}\left(v_{0}\right) \implies \operatorname{FUNCTION}\left(v_{0}\right)
\end{equation*}We then substitute ${{v_{64}}^{-1}}$ for $v_{0}$ to obtain 
\begin{equation} \label{17}
(\forall v_{64})\ {{v_{64}}^{-1}} \subseteq V \times V \;\wedge\; \operatorname{SINGLEVALUED}\left({{v_{64}}^{-1}}\right) \implies \operatorname{FUNCTION}\left({{v_{64}}^{-1}}\right)
\end{equation}We know the following: 
\begin{equation*} \label{8}
(\forall v_{0})\ {{v_{0}}^{-1}} \subseteq V \times V
\end{equation*}We then substitute $v_{64}$ for $v_{0}$ to obtain 
\begin{equation*} \label{18}
(\forall v_{64})\ {{v_{64}}^{-1}} \subseteq V \times V
\end{equation*}which together with (\ref{17}) yields 
\begin{equation*} \label{19}
(\forall v_{64})\ \operatorname{SINGLEVALUED}\left({{v_{64}}^{-1}}\right) \implies \operatorname{FUNCTION}\left({{v_{64}}^{-1}}\right)
\end{equation*}We then substitute $v_{0}$ for $v_{64}$ to obtain 
\begin{equation} \label{20}
(\forall v_{0})\ \operatorname{SINGLEVALUED}\left({{v_{0}}^{-1}}\right) \implies \operatorname{FUNCTION}\left({{v_{0}}^{-1}}\right)
\end{equation}We know the following: 
\begin{equation*} \label{10}
(\forall v_{0})\ \operatorname{SINGLEVALUED}\left(v_{0}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left(v_{0}\right),\operatorname{sv_{1}}\left(v_{0}\right)\right\rangle  \in v_{0}
\end{equation*}We then substitute ${{v_{0}}^{-1}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{21}
(\forall v_{0})\ \operatorname{SINGLEVALUED}\left({{v_{0}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{v_{0}}^{-1}}\right),\operatorname{sv_{1}}\left({{v_{0}}^{-1}}\right)\right\rangle  \in {{v_{0}}^{-1}}
\end{equation*}which together with (\ref{20}) yields 
\begin{equation} \label{22}
(\forall v_{0})\ \operatorname{FUNCTION}\left({{v_{0}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{v_{0}}^{-1}}\right),\operatorname{sv_{1}}\left({{v_{0}}^{-1}}\right)\right\rangle  \in {{v_{0}}^{-1}}
\end{equation}We know the following: 
\begin{equation*} \label{11}
(\forall v_{0})\ \operatorname{SINGLEVALUED}\left(v_{0}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left(v_{0}\right),\operatorname{sv_{2}}\left(v_{0}\right)\right\rangle  \in v_{0}
\end{equation*}We then substitute ${{v_{0}}^{-1}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{27}
(\forall v_{0})\ \operatorname{SINGLEVALUED}\left({{v_{0}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{v_{0}}^{-1}}\right),\operatorname{sv_{2}}\left({{v_{0}}^{-1}}\right)\right\rangle  \in {{v_{0}}^{-1}}
\end{equation*}which together with (\ref{20}) yields 
\begin{equation} \label{28}
(\forall v_{0})\ \operatorname{FUNCTION}\left({{v_{0}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{v_{0}}^{-1}}\right),\operatorname{sv_{2}}\left({{v_{0}}^{-1}}\right)\right\rangle  \in {{v_{0}}^{-1}}
\end{equation}In (\ref{16}) we substitute $\mathscr{P}(v_{64})$ for $v_{1}$ to obtain 
\begin{equation*} \label{29}
(\forall v_{0},v_{64})\ \left\langle v_{0},\mathscr{P}(v_{64})\right\rangle  \in \text{\emph{POWER}} \implies \operatorname{U}\left(\mathscr{P}(v_{64})\right) = v_{0}
\end{equation*}which together with $(\forall v_{64})\ \operatorname{U}\left(\mathscr{P}(v_{64})\right) = v_{64}$ yields 
\begin{equation*} \label{31}
(\forall v_{0},v_{64})\ \left\langle v_{0},\mathscr{P}(v_{64})\right\rangle  \in \text{\emph{POWER}} \implies v_{0} = v_{64}
\end{equation*}We then substitute $v_{1}$ for $v_{64}$ to obtain 
\begin{equation} \label{32}
(\forall v_{0},v_{1})\ \left\langle v_{0},\mathscr{P}(v_{1})\right\rangle  \in \text{\emph{POWER}} \implies v_{0} = v_{1}
\end{equation}We know the following: 
\begin{equation*} \label{2}
(\forall v_{0})\ \operatorname{FUNCTION}\left(v_{0}\right) \;\wedge\; \operatorname{FUNCTION}\left({{v_{0}}^{-1}}\right) \implies \operatorname{ONEONE}\left(v_{0}\right)
\end{equation*}We then substitute $\text{\emph{POWER}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{33}
\operatorname{FUNCTION}\left(\text{\emph{POWER}}\right) \;\wedge\; \operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right) \implies \operatorname{ONEONE}\left(\text{\emph{POWER}}\right)
\end{equation*}which together with $\operatorname{FUNCTION}\left(\text{\emph{POWER}}\right)$ yields 
\begin{equation} \label{34}
\operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right) \implies \operatorname{ONEONE}\left(\text{\emph{POWER}}\right)
\end{equation}In (\ref{22}) we substitute $\text{\emph{POWER}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{35}
\operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation*}which together with (\ref{34}) yields 
\begin{equation*} \label{36}
\operatorname{ONEONE}\left(\text{\emph{POWER}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation*}which together with $\neg \operatorname{ONEONE}\left(\text{\emph{POWER}}\right)$ yields 
\begin{equation} \label{37}
\left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation}We know the following: 
\begin{equation} \label{3}
(\forall v_{0},v_{1},v_{2})\ \left\langle v_{0},v_{1}\right\rangle  \in {{v_{2}}^{-1}} \implies \left\langle v_{1},v_{0}\right\rangle  \in v_{2}
\end{equation}We then substitute $\operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{0}$ and $\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{1}$ and $\text{\emph{POWER}}$ for $v_{2}$ to obtain 
\begin{equation*} \label{38}
\left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}} \implies \left\langle \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in \text{\emph{POWER}}
\end{equation*}which together with (\ref{37}) yields 
\begin{equation} \label{39}
\left\langle \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in \text{\emph{POWER}}
\end{equation}In (\ref{28}) we substitute $\text{\emph{POWER}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{42}
\operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation*}which together with (\ref{34}) yields 
\begin{equation*} \label{43}
\operatorname{ONEONE}\left(\text{\emph{POWER}}\right) \;\vee\; \left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation*}which together with $\neg \operatorname{ONEONE}\left(\text{\emph{POWER}}\right)$ yields 
\begin{equation} \label{44}
\left\langle \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation}In (\ref{6}) we substitute $\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{0}$ and $\operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{1}$ to obtain 
\begin{equation*} \label{45}
\left\langle \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right),\operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in \text{\emph{POWER}} \implies \mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)) = \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with (\ref{39}) yields 
\begin{equation*} \label{46}
\mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)) = \operatorname{sv_{3}}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with (\ref{44}) yields 
\begin{equation} \label{48}
\left\langle \mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)),\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}}
\end{equation}In (\ref{3}) we substitute $\mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right))$ for $v_{0}$ and $\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{1}$ and $\text{\emph{POWER}}$ for $v_{2}$ to obtain 
\begin{equation*} \label{49}
\left\langle \mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)),\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)\right\rangle  \in {{\text{\emph{POWER}}}^{-1}} \implies \left\langle \operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right),\mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right))\right\rangle  \in \text{\emph{POWER}}
\end{equation*}which together with (\ref{48}) yields 
\begin{equation} \label{50}
\left\langle \operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right),\mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right))\right\rangle  \in \text{\emph{POWER}}
\end{equation}In (\ref{32}) we substitute $\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{0}$ and $\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)$ for $v_{1}$ to obtain 
\begin{equation*} \label{51}
\left\langle \operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right),\mathscr{P}(\operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right))\right\rangle  \in \text{\emph{POWER}} \implies \operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right) = \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with (\ref{50}) yields 
\begin{equation} \label{52}
\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right) = \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation}In (\ref{4}) we substitute ${{\text{\emph{POWER}}}^{-1}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{53}
\operatorname{sv_{2}}\left({{\text{\emph{POWER}}}^{-1}}\right) = \operatorname{sv_{1}}\left({{\text{\emph{POWER}}}^{-1}}\right) \implies \operatorname{SINGLEVALUED}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with (\ref{52}) yields 
\begin{equation*} \label{54}
\operatorname{SINGLEVALUED}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}In (\ref{20}) we substitute $\text{\emph{POWER}}$ for $v_{0}$ to obtain 
\begin{equation*} \label{59}
\operatorname{SINGLEVALUED}\left({{\text{\emph{POWER}}}^{-1}}\right) \implies \operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with $\operatorname{SINGLEVALUED}\left({{\text{\emph{POWER}}}^{-1}}\right)$ yields 
\begin{equation*} \label{60}
\operatorname{FUNCTION}\left({{\text{\emph{POWER}}}^{-1}}\right)
\end{equation*}which together with (\ref{34}) yields 
\begin{equation*} \label{63}
\operatorname{ONEONE}\left(\text{\emph{POWER}}\right)
\end{equation*}but this contradicts $\neg \operatorname{ONEONE}\left(\text{\emph{POWER}}\right)$.

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