$\Pr(A \cap B \cap C)=\Pr(A\cap B \cap D)$ implies $\Pr(A \cap B \cap C^c)=\Pr(A\cap B \cap D^c)$? Consider four events $A,B,C,D$. Assume that
$$
\Pr(A \cap B \cap C)=\Pr(A\cap B \cap D)
$$
Is it true that
$$
\Pr(A \cap B \cap C^c)=\Pr(A\cap B \cap D^c)
$$
where $C^c$ and $D^c$ are the complements of $C$ and $D$, respectively?

The expression can be further simplified by setting $E = A\cap B$. Thus your question becomes:
Does (the premise)
$$P(E\cap D)$$ imply
$$
P(E\cap C^c) = P(E\cap D^c)?
$$
Proof. Yes indeed,
$$P(E\cap C^c) = P(E) - P(E\cap C) ,\tag{*}$$
and
$$P(E)-E(E\cap D) = P(E\cap D^c).\tag{**}$$
Using the premise in (*) and equating this with (**), we get the desired equality.
