$\Pr(A \cap B \cap C)=\Pr(A\cap B \cap D)$ 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?

Yes, just note
\begin{align}
& P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C^c), \\
& P(A \cap B) = P(A \cap B \cap D) + P(A \cap B \cap D^c).
\end{align}
