How to prove if $A \subseteq B$ then $B \setminus (B \setminus A) = A$ using forward-backward method?

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?

• I am unfamiliar with the book you are using, but in my (limited) experience it is frowned upon to use the "$\in$" symbol to denote a subset. – eepperly16 Mar 2 '16 at 3:52
• In general, when referencing a book, you should mention title and author, maybe edition, not publisher. Dover in particular republished many old books. Also, make sure (as you did) that when referencing a book, you either re-state the relevant parts in the question (e.g. the problem), or provide a link to the book where someone could read it (i.e. don't send us to Amazon, but a PDF of the book; nobody is buying a textbook to answer your question). – AJY Mar 2 '16 at 5:46
• You dont need quantifiers. It is obvious that if $x\in A$ then $x\in B$, no matter what $x$. – Masacroso Mar 2 '16 at 6:07

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 $A$.
• @BrendanHill The above is completely rigorous, albeit not formal. This method you're talking about probably means that to show that two sets $T,S$ are equal, it suffices to show $S$ is contained in $T$ and conversely. I did that, but using words. You'll profit more from being a clear user of words than from jamming symbols into paper! – Pedro Tamaroff Mar 2 '16 at 4:02