From "Introduction to Real Analysis (Dover books)" Exercise 1.15.1.g is to prove:
if $A \subseteq B$ then $B \setminus (B \setminus A) = A$
It indicates to use the forward backwards method. I an struggling to fill in the middle and unsure if I have formulated the conclusion correctly. Note that my comments for the backwards part relate to how the upper statement is derived from the lower statement.
$$A \subseteq B$$ $$(\forall x)(x \in A \implies x \in B) $$ $$\vdots$$
$$(\forall x)(((x \in B) \land (x \in A)) \iff (x \in A)) $$
Removing the contradictory disjunct, removing double negation ^^
$$(\forall x)((((x \in B) \land \neg(x \in B)) \lor ((x \in B) \land (\neg(\neg(x \in A))) \iff (x \in A)) $$
Distributing the disjunct over the conjunct ^^
$$(\forall x)(((x \in B) \land (\neg(x\in B) \lor \neg(\neg(x \in A))) \iff (x \in A)) $$
Processing the negated conjunction ^^
$$(\forall x)(((x \in B) \land\neg((x\in B) \land(\neg(x \in A))) \iff (x \in A)) $$ Conclusion: $$B \setminus (B \setminus A) = A$$
It's not clear to me how to get from the premise to the the biconditional statement?
Also I wonder if there is a much simpler proof?
Also have I formulated $B \setminus (B \setminus A) = A$ correctly?
You simply need to unwrap the definitions. For a subset $S$ of $B$, the set $B\setminus S$ is the collection of $x\in B$ that do not lie in $S$. Setting $B\setminus A=S$, we get $B\setminus (B\setminus A)$ is the collection of $x\in B$ that do not lie in $B\setminus A$. But an element of $B$ lies in $B\setminus A$ if and only if it lies in $B$ but not in $A$. Thus $B\setminus (B\setminus A)$ is the set of elemets of $B$ for which it is not true they do not lie in $A$. This is just the set of elements that do lie in $A$, which is $A$.