If A,B and C are mutually independent, how to prove that $A'$ and $B \cap C'$, $B'$ and $A \cap C'$, $A'$ and $C \cap B'$ are mutually independent? If A,B and C are mutually independent, how to prove that $A'$ and $B \cap C'$,  $B'$ and $A \cap C'$, $A'$ and $C \cap B'$ are mutually independent? Any properties and definitions to consider?
 A: Use the simple property:
$$
(X\cap Y) \cup (X\cap Y') = X, \quad X\cap Y \; \text{ and } \; X\cap Y' \; \text{ are disjoint}
$$
which implies 
$$\tag{1}
\mathbb P(X\cap Y')=\mathbb P(X)-\mathbb P(X\cap Y).
$$
To prove $A'$ and $B\cap C'$ are independent, consider 
$$
\mathbb P(A'\cap B\cap C')\; \overset{(1)}{=} \; \mathbb P(A'\cap B) - \mathbb P(A'\cap B \cap C) 
$$
apply (1) again
$$
=\left(\mathbb P(B) - \mathbb P(A\cap B)\right) - \left(\mathbb P(B \cap C) - \mathbb P(A \cap B\cap C)\right) 
$$
$$
=\mathbb P(B) - \mathbb P(A)\mathbb P(B) - \mathbb P(B)\mathbb P(C) + \mathbb P(A)\mathbb P(B)\mathbb P(C) \tag{2}
$$
Compare it with product of probabilities:
$$
\mathbb P(A')=1-\mathbb P(A), 
$$
$$\mathbb P(B\cap C') = \mathbb P(B) - \mathbb P(B\cap C) = \mathbb P(B) - \mathbb P(B)\mathbb P(C)
$$
$$
\mathbb P(A')\mathbb P(B\cap C') = (1-\mathbb P(A))\cdot (\mathbb P(B) - \mathbb P(B)\mathbb P(C)) 
$$
$$ = \mathbb P(B) - \mathbb P(A)\mathbb P(B) - \mathbb P(B)\mathbb P(C) + \mathbb P(A)\mathbb P(B)\mathbb P(C) \tag{3}
$$
Comparing (2) and (3) conclude independence. 
A: $P(B\cap C')=P(B)-P(B\cap C)(=P(B)-P(B)P(C))$ because $B\cap C'=B \setminus B\cap C$ (and $B\cap C$ is  a subset of $B$). Now $P(A' \cap (B\cap C'))=P(B\cap C')-P(A\cap B\cap C')$ for a similar reason. Also $P(A\cap B\cap C')=P(A \cap B) -P(A\cap B \cap C)=P(A) P(B) -P(A)P(B)P(C)$. Combine all these to get $P(A' \cap (B\cap C'))=P(A') P(B \cap C')$. The other two facts are proved similarly. 
