In probability, if we have an equality, can we condition on an arbitrary event and guarantee that the equality still holds? I have a question regarding equalities and conditioning probabilities. I was reading through this proof for conditional independence, which can be found here. In the proof, they first state that $p(A|B)=\frac{p(A\cap B)}{P(B)}$, which comes from the definition of conditional probability. This I understand. But, they then go on to state that $p(A|B,C)=\frac{p(A\cap B|C)}{P(B|C)}$ by conditioning on $C$. This seems like a strong claim. Is it is case that when you have an equality of probabilities, you may condition on any arbitrary event given that it occurs with a non-zero probability? For example, if I know that $p(A\cap B) + p(A^c\cap B)=p(B)$, may I also state that $p(A\cap B | C) + p(A^c\cap B | C)=p(B | C)$ by conditioning on $C$?
Further, if this is true, how might I go about proving it? And if it's false, is there a counter-example?

If you understand that $P(A|B, C)$ really means $P(A|B \cap C)$ (in probability, comma "," is frequently interpreted as the synonym for "intersection"), then it is still a matter of applying definition:
\begin{align}
 & P(A|B \cap C) \\
=& \frac{P(A \cap B \cap C)}{P(B \cap C)} \\
=& \frac{P(A \cap B | C) P(C)}{P(B | C)P(C)} \\
=& \frac{P(A \cap B | C)}{P(B | C)}.
\end{align}
This completes the proof.
