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\begin{definition}\label{phase5_replacement_definition}
$\phasefivereplacement{A} = \{ y \mid x \in A, y \in x \mid x = y \}$.
\end{definition}
\begin{proposition}\label{phase5_replacement_theorem}
For all $A, x$ we have if $x \in A$, then $x \in \{ y \mid y \in A \}$.
\end{proposition}
\begin{proof}
Fix $A, x$.
Assume $x \in A$.
\end{proof}
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