How to introduce bi-conditional in this proof?

Question

This is from Discrete Mathematics and its Applications

Just for context, I know that the universal set is everything and that the complement of A is just difference of the universal set and A. A good diagram

Here is my work so far

When I first saw the world if and only if, I know that involves the bi-conditional(imply one from the other). Via my work I have shown that if A is a subset of B, the complement of B is a subset of the complement of A. To prove the bi-condition, would I have to just copy and paste/ reverse my initial proof or is there some shortcut you can take. To me, writing the whole thing out again seemed redundant.

Answer 1

You can avoid "copy-past" it iff you note that the key-point in the proof uses contraposition, i.e. :

$(p \rightarrow q) \leftrightarrow (\lnot q \rightarrow \lnot p)$.

This means that in your proof you have used it to pass from :

$x \in A \rightarrow x \in B$

to :

$\lnot (x \in B) \rightarrow \lnot (x \in A)$ i.e. $(x \notin B \rightarrow x \notin A)$.

For the other direction, when you "reverse" the steps of the first part of the proof, you have to use the other side of the bi-implication, passing from :

$x \notin B \rightarrow x \notin A$ i.e. $\lnot (x \in B) \rightarrow \lnot (x \in A)$

to :

$x \in A \rightarrow x \in B$.