Proof that $P(A \mid C) = 1$ implies $P(A \mid B∩ C) = 1$ This isn't homework. I'm reading Probabilistic Graphical Models by Koller et al.$\space$, and an easy problem in chapter $3$ made me think of a more general problem (which I'm now stuck on). I have everything in place except for this:

Let $P(B \cap C) \neq 0$. Then $P(A \space|\space B ∩ C) = 1$ if $P(A \mid C) = 1$. It's intuitively obvious, but I haven't been able to formally prove it. I get

$$P(A \mid B∩C) = \dfrac{P(A∩B∩C)}{P(B∩C)} = \dfrac{P(C) P(A\mid C) P(B\mid A∩C)}{P(B∩C)} = \dfrac{P(B\mid A∩C)}{P(B|C)}$$

I don't see why that ratio on the right hand side is $1$. Is it? What am I not seeing?
Assume that $\mathbb P(\ \mid C)$ and $\mathbb P(\ \mid B\cap C)$ both exist, that is, that 
$$\mathbb P(B\cap C)\ne0.
$$
Then note that $\mathbb P(A\mid C)=1$ if and only if $\mathbb P(A\cap C)=\mathbb P(C)$ if and only if $\mathbb P(C\setminus A)=0$. Likewise, $\mathbb P(A\mid B\cap C)=1$ if and only if $\mathbb P((B\cap C)\setminus A)=0$. But $(B\cap C)\setminus A\subseteq C\setminus A$, hence the latter implies the former, that is,
$$
\mathbb P(A\mid C)=1\implies \mathbb P(A\mid B\cap C)=1.
$$
$\mathbb{P}\{A|C\}=1$ implies $\mathbb{P}\{A,C\}=\mathbb{P}\{C\}$, which implies $A\subset C$. Therefore $\mathbb{P}\{A|B,C\}=1$.