I'm trying to proof that $P\left[(A|B)\,\cap C\right] = P\left[A|\,(B\cap C)\right]$ using the conditional probability definition: $$ P(X|Y)=\frac{P(X\cap Y)}{P(Y)} \implies P(X\cap Y)=P(X|Y) \cdot P(Y) $$ Expanding the left side of the equation using $P(X\cap Y)=P(X|Y) \cdot P(Y)$, we have: $$ P\left[(A|B)\,\cap C\right] =P[(A|B)\,|C]\cdot P(C) $$ I don't know how to expand $P[(A|B)\,|C]$ and how to proceed to prove the equality. How can I prove it?
Context:
I was trying to solve a conditional probability problem in which one of the given values was $P[(A|B) \cap C]$. The problem asked for $P(A\cap B \cap C)$.
So, to solve it, I used: $$ P(A \cap B \cap C) = P(C)\cdot P(B|C) \cdot P(A|B \cap C) $$
And that's why I'm trying to prove that $P\left[(A|B)\,\cap C\right] = P\left[A|\,(B\cap C)\right]$: to use the data given by the problem in the formula above.