TITLE: Is proving "not $A \iff$ not $B$" the same as proving $A \iff B$?
QUESTION [5 upvotes]: Is proving "not $A \iff$ not $B$" the same as proving $A \iff B$?
I'm wondering because I'm looking at a statement in my book of the form $A \iff B$ yet the proof shows not $A \iff$ not $B$ and then says it is complete.
So I think they would be the same but I don't exactly see why.
Anyone explain to me please?

REPLY [3 votes]: I'll assume you know the contrapositive: proving $p\rightarrow q$ is the same as proving $\neg q \rightarrow \neg p$.
With this, it's easy:
$$(p\iff q) \equiv ((p\rightarrow q)\wedge (q\rightarrow p))\equiv ((\neg q\rightarrow \neg p)\wedge (\neg p\rightarrow \neg q))\equiv (\neg q \iff \neg p)$$

REPLY [2 votes]: Think of it this way: $A\iff B$ is two statements: $A\implies B$ and $B\implies A$.
Using prime to denote "not", by contraposition we have $B'\implies A'$ and $A'\implies B'$. Thus $A'\iff B'$.
Since this proof goes the other way, we have $(A\iff B)\iff (A'\iff B')$