Proving Set Identities - Why Do I Need Both Directions? Say I want to prove $A - B = A \cap B'$
I let $x \in A - B$
Then $x \in A$ and $x \notin B$
By definition, $x \notin B \iff x \in B'$ (no?)
So the thing looks proven to me.
Why do I need to prove that $A - B \subseteq A \cap B'$ and $A \cap B' \subseteq A - B$ ?
More generally is there a class of set theory proofs which require set equality via mutual subset inclusion and another class which can be proven directly? If so, how do I decide which technique to use please?
 A: "$X=Y$" in set theory means that any formula ("property") that the set $X$ satisfies is also satisfied by the set $Y$, and vice versa. This is, in somewhat informal terms, the definition of $=$. Given a specific element $x$, the property $\phi(Z)$ given by $x\in Z$ is one such property, so by definition of $=$, equal sets contain all the same elements. But there are many other formulas one could ask a set to fulfill as well.
There is an axiom of set theory (at least in ZF set theory, the most used set theory in modern mathematics) which says that when you want to check whether two sets are equal, you don't have to check every formula to see that $X$ and $Y$ either both satisfy the formula, or they both do not satisfy it. The axiom states that if they contain the same elements, they are proclaimed to be equal (in the sense of the first paragraph).
Thus the minimal thing you need to show in order to decide that $X = Y$, according to the axioms of set theory, is exactly that
$$
x\in X\implies x\in Y\\
y\in Y \implies y\in X
$$
So yes, formally, you do have to go both ways.
Of course, with simple, bare-bones examples like yours, we do get that the proof of one implication looks a whole lot like the proof of the other implication in revese, and with only two or three steps to the proof, it's hard to know where the proof of one implication ends and the proof of the other implication begins if you're not very careful to keep the two apart.
