simple probability identity I got a little stuck on a simple proof of the following probability identity. 
Given
$\mathbb{P}(A^c \cap B^c)=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\cap B)$
how to prove for any set $X$,
$\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B)$
Looks very intuitive; just replace the whole space by $X$. But how to prove it simply and rigorously? Thanks.
 A: Write $X$ as a disjoint union:
$$
X=(X\cap A\cap B)\sqcup (X\cap A^c\cap B)\sqcup (X\cap A\cap B^c)\sqcup (X\cap A^c\cap B^c).
$$
This gives:
$$
P(X)=P(X\cap A\cap B)+P(X\cap A^c\cap B)+P(X\cap A\cap B^c)+P(X\cap A^c\cap B^c).
$$
Hence
$$
P(X\cap A^c\cap B^c)=P(X)-P(X\cap A\cap B)-P(X\cap A^c\cap B)-P(X\cap A\cap B^c).
$$
Now the result follows from $P(X\cap A\cap B)+P(X\cap A^c\cap B)=P(X\cap B)$ and $P(X\cap A\cap B)+P(X\cap A\cap B^c)=P(X\cap A)$.
A: An easier way would be:
$X \cap A^c \cap B^c = X - (A \cup B)$.
Therefore,  $P(X - (A \cup B)) \\= P(X) - P(X \cap (A\cup B)) \\= P(X) - P((X\cap A)\cup(X\cap B)) \\= P(X) - [P(X\cap A) + P(X\cap B) - P(X\cap A\cap B)]$
A: Just follow your intuition: if $\mathbb{P}(X) > 0$ the formula $\mathbb{P}_X(E) = \dfrac{\mathbb{P}(E \cap X)}{\mathbb{P}(X)}$ defines a new (conditionnal) probability on $\Omega$. For this probability, you know that:
$$\mathbb{P}_X(A^c \cap B^c)=1-\mathbb{P}_X(A)-\mathbb{P}_X(B)+\mathbb{P}_X(A\cap B).
$$
Multiplying by $\mathbb{P}(X)$ gives what you want:
$$
\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B).
$$
Notice that the identity is trivial when $\mathbb{P}(X)=0$, since every term is $0$.
A: so.. i am using the latter to derive the former..


*

*write P(X)  as $\mathbb{P}(X\cap (A U A^c))$ , equal to $P(X \cap A)+P(X \cap A^c)$, cancel some terms

*then expand $P(X \cap A^c)$ to $P(X \cap A^c \cap (B U B^c))$, which is equal to $P(X \cap A^c \cap B)+ P(X\cap A^c \cap B^c)$, cancel the terms to get the answer.
