\begin{proposition}\label{bounded_fix_single}
    Let $A$ be a set.
    For all $x\in A$ we have $x\in A$.
\end{proposition}
\begin{proof}
    Fix $x\in A$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{bounded_fix_multiple}
    Let $A$ be a set.
    For all $x,y\in A$ we have $x\in A$ and $y\in A$.
\end{proposition}
\begin{proof}
    Fix $x,y\in A$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{bounded_fix_negative}
    Let $A$ be a set.
    For all $x\notin A$ we have $x\notin A$.
\end{proposition}
\begin{proof}
    Fix $x\notin A$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{fix_such_that}
    Let $A$ be a set.
    For all $x$ such that $x\in A$ we have $x\in A$.
\end{proposition}
\begin{proof}
    Fix $x$ such that $x\in A$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{assume_left_conjunct}
    Let $A,B$ be sets.
    If $A=A$ and $B=B$, then $A=A$.
\end{proposition}
\begin{proof}
    Assume $A=A$.
    Assume $B=B$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{assume_right_conjunct}
    Let $A,B$ be sets.
    If $A=A$ and $B=B$, then $B=B$.
\end{proposition}
\begin{proof}
    Assume $B=B$.
    Assume $A=A$.
    Follows by assumption.
\end{proof}

\begin{proposition}\label{take_bounded}
    Let $A$ be a set.
    Suppose there exists $x\in A$ such that $x=x$.
    Then $A=A$.
\end{proposition}
\begin{proof}
    Take $x\in A$ such that $x=x$ by assumption.
    We have $x\in A$ by assumption.
    Follows.
\end{proof}

\begin{proposition}\label{take_named_noun}
    Let $A$ be a set.
    Suppose there exist sets $x,y$ such that $x=x$ and $y=y$.
    Then $A=A$.
\end{proposition}
\begin{proof}
    Take a set $x,y$ such that $x=x$ and $y=y$ by assumption.
    Follows.
\end{proof}

\begin{proposition}\label{take_anonymous_noun}
    Let $A$ be a set.
    Suppose there exists a set.
    Then $A=A$.
\end{proposition}
\begin{proof}
    Take a set by assumption.
    Follows.
\end{proof}

\begin{proposition}\label{existential_have_witness}
    Let $A$ be a set.
    Suppose there exists $x\in A$ such that $x=x$.
    Then $A=A$.
\end{proposition}
\begin{proof}
    We have there exists $x\in A$ such that $x=x$ by assumption.
    We have $x\in A$ by assumption.
    Follows.
\end{proof}

\begin{proposition}\label{take_omitted_continuation}
    Let $A$ be a set.
    Suppose there exists $x\in A$ such that $x=x$.
    Then $A=A$.
\end{proposition}
\begin{proof}
    Take $x\in A$ such that $x=x$ by assumption.
    Omitted.
\end{proof}
