= Solution
The first condition concerns the domains: their <symmetric difference> $A\mathbin\triangle B$ must be a <finite set>. The original PDF prints this condition and the agreement condition together; the converted TeX omits both. The two existing solution scopes are used here to prove the domain and value parts of the single <equivalence relation> assertion.
Domain agreement is a <reflexive relation> because $A\mathbin\triangle A=\varnothing$, and a <symmetric relation> because $A\mathbin\triangle B=B\mathbin\triangle A$. For the <transitive relation> property, membership can differ between $A$ and $C$ only if it differs between $A$ and $B$ or between $B$ and $C$. Thus
$$
\boxed{A\mathbin\triangle C\subseteq(A\mathbin\triangle B)\cup(B\mathbin\triangle C).}
$$
A <set union> of two <finite sets> is finite, so the domain condition passes through a chain of related pairs. This is the domain part of <equivalence of partial functions modulo finite changes>.
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