Example to show that $P(A - B)$ need not equal $P(A) - P(B)$ if $B$ is not a subset of $A$? I am trying to understand the example given in the text (Basic Probability Theory, Robert Ash; Section 1.3 #2, https://faculty.math.illinois.edu/~r-ash/BPT.html) for above statement. Below is the example given in the "solutions to problems not in text": 

$P(A) = P(A - B) + P(A \cap B)$, so any example in which $P(A \cap B) < P(B)$ will do (e.g., let $A = B^\complement$)

There are several things I don't understand in the solution. 
Firstly, how did we arrive at the definition that $P(A) = P(A - B) + P(A \cap B)$? Of course, if I draw a venn diagram where A and B are not disjoint sets, it makes sense. Is that correct? How do we arrive at that formula if we want to use mathematical approach?
Secondly, how did we arrive at the conclusion that any with $P(A \cap B) < P(B)$ would work? 
Thanks for any pointers!
 A: Note that $$(A \cap B^c) \cup (A \cap B)=A$$
In english, it says, an element of $A$ is either in $B$ or not in $B$.
Also, we have $(A \cap B^c) \cap (A \cap B) = \emptyset$ since $B$ and $B^c$ are disjoint. 
Hence $$P(A \cap B^c) + P(A \cap B) = P(A)$$
$$P(A -B) + P(A \cap B) = P(A)$$
$$P(A -B)  = P(A)-  P(A \cap B)$$
If $P(A \cap B) < P(B)$, then 
$$-P(A \cap B) > -P(B)$$
$$P(A)-P(A \cap B) > P(A)-P(B)$$
$$P(A-B) > P(A)-P(B)$$
A: Take a non-empty set $X$ and a Boolean algebra $\Bbb B$ of subsets of $X.$ That is,
(i). $X\in \Bbb B.$
(ii). $\forall b\in \Bbb B\,(X- b\in \Bbb B).$
(iii).$\forall b_1,b_2\in \Bbb B\;(b_1\cup b_2\in\Bbb B).$
It follows that $\emptyset \in \Bbb B$ and that both $\cap C$ and $\cup C$ belong to $\Bbb B$ whenever $C$ is a finite subset of $B.$ And that $A-B\in \Bbb B$ when $A,B\in \Bbb B. $
$\Bbb B$ may be the set $\Bbb P(X)$ of all subsets of $X $ or we may have $B=\{X, \emptyset\}$ or $\Bbb B$ may be "in between" these two extremes.
Let $P:\Bbb B\to [0,1]$ be a function such that $P(X)=1$ 
 and such that $P(b_1\cup b_2)=P(b_1)+P(b_2)$ when $b_1,b_2$ are disjoint members of $\Bbb B.$
The triplet $(X,\Bbb B, P)$ is called a Probability Space.
Now $A-B$ and $A\cap B$ are disjoint, and their union is $A,$  so if $A,B\in \Bbb B$ then $P(A)=P(\,(A-B)\cup (A\cap B)\,)=$ $=P(A-B)+P(A\cap B).$
Colloquially your chance of winning the lottery today is equal to your chance of winning the lottery today and not  dying today PLUS your chance of winning the lottery today and  dying today.
In many cases we would also like $\cup_{n\in \Bbb N}b_n \in \Bbb B$ whenever $\{b_n:n\in \Bbb N\}\subset B,$ in which case $\Bbb B$ is called a $\sigma$-algebra. And usually in such  a case , we  also want $P$ to be countably additive, which means that $P(\cup_{n\in \Bbb N}b_n)=\sum_{n\in \Bbb N} P(b_n),$ if $b_m,b_n$ are disjoint members of $\Bbb B$ whenever $m \ne n.$
