= Solution
\b[True.] The two operations retain complementary membership information. For $x\in C$, equality of the <intersections> makes $x\in A$ equivalent to $x\in B$. For $x\notin C$, equality of the <unions> makes $x\in A$ equivalent to $x\in B$. These cases exhaust every possible element. <Set extensionality> therefore gives $A=B$. Equivalently, reconstruct either <set> from
$$
A=(A\cap C)\cup\bigl((A\cup C)\setminus C\bigr).
$$
Substituting the two given equalities into this expression gives $B$.
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