Proof of $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ Reading a book where proof which I mentioned is described as follows :
$$\ldots=P((A\cap B’)\cup(A\cap B)) + P((A'\cap B)\cup(A\cap B)) - P(A\cap B) = P(A) + P(B) - P(A\cap B)$$
What I don't understand here why does
$$P((A\cap B’)\cup(A\cap B))$$ equals to
$$P(A)$$
and
$$P((A'\cap B)\cup(A\cap B))$$
$$P(B)$$
respectively?
 A: If an element belongs to $A$ but not to $B$ or both to $A$ and $B$, then it belongs to $A$. This is elementary set algebra.
A: $$((∩′)∪(∩))$$
and $$P(A)$$ are equal for a basic reason: $$(∩′)∪(∩)=A.$$
The two sets contain exactly the same events.
Let's say that $A$ is the set of events where I have asparagus for lunch.
Let's say that $B$ is the set of events where it's raining.  Then $B'$ is the set of events where it's not raining.
Suppose I have asparagus.   We could ask if it is raining or not.
$A\cap B'$ is the events where I have asparagus and it is not raining.  $A\cap B$ is the events where I have asparagus and it is raining.
$(A\cap B')\cup (A\cap B)$ is the events where I have asparagus and it is not raining, or where I have asparagus and it is raining.
But it either is raining, or it isn't.  There is no third possibility.  That is what $B'$ means.  It covers exactly the events that $B$ does not.
So $(A\cap B)\cup (A\cap B')$ is just the same a $A$, but split into two parts, one where it is raining and one where it isn't.
Since the two sets of events are the same set, they must have the same probability.
