How to prove / disprove these conditional probability statements? I was browsing a well known technology board and someone posted a pretty standard question about conditional probabilities. As this happens to be the current subject of my lectures (just starting), I decided to take a swing at them and it turns out I have no idea what I'm doing, or what kind of proof (contradiction?) I should be using in these kinds of questions. I can show these are correct/incorrect with arbitrary numbers, but I have a gut feeling there are more succinct ways of showing truth.

If P(a|b,c) = P(a), then P(b|c) = P(b)

I get the feeling that this one is wrong, only because we can't assume c has no bearing on b with only the information that neither affects a. From my limited knowledge of conditional probability, I'm reading the first statement that a is independent from both b and c. I did this, where I assumed the implication was true, and got a contradiction. Nevermind, I made the assumption that P(a|b,c) = P(a) ==> P(a|b) = P(a) and P(a|c) = P(a), and I apparently can't do that?

If P(a | b) = P(a), then P(a | b, c) = P(a | c)

I believe this one is true, because if a and b are independent, then the probability of a, given two conditions - one of which doesn't affect it, and one that might - should just be the probability of a given the other condition (c). I also tried proving this one, but I kept going in a circle.
I've tried contradiction approaches, but I'm not familiar enough with conditional probability to say something is for sure a contradiction (For example, before I realized the incorrect assumption I made on the first one, I assumed the second part was true and arrived at P(b|a,c) = P(b|c), which looks false, but because of the stuff about independence I can't be certain). Most of my 'technique' has revolved around using the definitions  

P(a|b) = P(a,b)/p(b) (expanded to) P(a|b,c) = P(a,b,c)/P(b,c)
  P(a,b) = P(a|b)P(b) = P(b|a)P(a)

Any help is appreciated, especially if you focus on the "why" of choosing a certain proof.
 A: \begin{equation}
\begin{split}
   P(a|b,c) &= P(a)\hspace{30pt}\text{(It means that $a$ is independent of $b$ and $c$)}\\
\Rightarrow \frac{P(a,b,c)}{P(b,c)}&=P(a)\\
\Rightarrow P(a,b,c)&=P(a)P(b,c) = P(a)P(b|c)P(c)\\
\Rightarrow P(a,b,c)&=P(a)P(c)P(b|c)
\end{split}
\end{equation}
If $P(b|c) = P(b)$, then 
\begin{equation}
\begin{split}
   P(a,b,c)&=P(a)P(c)P(b)
\end{split}
\end{equation}
which means that $a,b,c$ are independent of each other, which means that $b$ is independent of $c$.
A: 

*

*$P(A|B\cap C) = P(A)\implies P(B|C) = P(B)$

counterexample:

$$P(A|B\cap C) \\= \frac{P(ABC)}{P(ABC,A'BC)} \\= \frac12 \\= P(A),$$ but $$P(B|C)\\= \frac{P(ABC,A'BC)}{P(ABC,AB'C,A'BC,A'B'C)} \\= \frac13 \\\neq \frac12 \\= P(ABC,ABC',A'BC,A'BC') \\= P(B).$$




*$P(A|B) = P(A)\implies P(A|B\cap C) = P(A|C)$

counterexample:
One ball is randomly drawn from a bowl containing four balls numbered 1, 2, 3, and 4.
Define event $A = \{1,2\},$ event $B = \{1,3\},$ event $C = \{1,4\}.$
Then $P(A)=P(B)=P(C)=\frac12,$ $P(A\cap B)=P(B\cap C)=P(C\cap A)=\frac14,$ and $P(A\cap B \cap C)=\frac14.$
$$P(A|B) \\= \frac14\div\frac12 \\= \frac12\\=P(A),$$ but $$P(A|B\cap C)\\= \frac14\div\frac14 \\\neq \frac14\div\frac12 \\= P(A|C).$$
