Does $p(a | b,c) = p(a|c)$ necessarily imply $p(a|b) = p(a)$?

Question

I'm in a ML course, and we had this math refresher quiz. We were asked to prove (or disprove) the following: $$p(a | b,c) = p(a|c) \to p(a|b) = p(a).$$

It is clear that $a$ is not dependent on $b$, when both $b$ and $c$ occur. However, it is not fully clear wether this implies that $a$ itself is independent of $b$. Intuitively, yes. But I don't know how to prove that mathematically. Any suggestions?

Answer 1

$$p(a|b,c)=\frac{p(a,b,c)}{p(b,c)}=\frac{p(a,c)}{p(c)}\frac{p(b|a,c)}{p(b|c)}=$$

$$p(a|b,c)=p(a|c)\cdot P$$

$$P=\frac{p(a,b,c)}{p(a,c)}\cdot\frac{p(c)}{p(b,c)}$$

Now consider the following sets

... it is self evident that $P=1$ and thus

$p(a|b,c)=p(a|c)$ but $a,b$ are not independent as $b \subset a$

In other words, $p(a|b,c)=p(a|c)$ is NOT SUFFICIENT for independence between $a$ and $b$

Answer 2

Hint: Consider events $a$, $b$ and $c$ such that $p(b|\,c)=1$.