 Solution

For the value condition, a function agrees with itself outside the empty exceptional set, proving the reflexive relation property. Interchanging the two functions preserves equality, so the same exceptional finite set proves the symmetric relation property.

Suppose $(A,f)$ is related to $(B,g)$ and $(B,g)$ to $(C,h)$, with exceptional finite sets $F_{AB}$ and $F_{BC}$. Points of $A\cap C$ might be absent from $B$, so merely taking $F_{AB}\cup F_{BC}$ would miss a real issue. Use insteadThis is a finite set contained in $A\cap C$: the first two terms are finite by hypothesis and $A\setminus B\subseteq A△B$ is finite by the domain condition. If $x\in (A\cap C)\setminus F_{AC}$, then $x\in B$ and neither original exception occurs, so $f(x)=g(x)=h(x)$. The domain condition is also preserved, as proved in the preceding scope. Thus the full relation is a transitive relation, andFinite domains cause no problem: one may take the whole intersection as the exceptional set. Both conditions are needed for equivalence of partial functions modulo finite changes.