(A iff B), then (not A iff not B)? Suppose $A\Leftrightarrow B$, i.e. statement A holds if and only if statement B holds.

Does then $\neg A\Leftrightarrow\neg B$?



I think this is true since
$A\implies B$ implies $\neg B\implies \neg A$

and

$B\implies A$ implies $\neg A\implies\neg B$.

Hence, if $A\Leftrightarrow B$, then
$$
\neg A\implies\neg B\implies\neg A
$$
which means that 
$$
\neg A\Leftrightarrow\neg B.
$$

I think, we even have
$$
(A\Leftrightarrow B)~\Leftrightarrow~(\neg A\Leftrightarrow\neg B)
$$
The symbol $\iff$ means if inputs are both true or false. So if A and b = 0 or 1, then A $\iff$ B is true otherwise false.

The $\lnot$ means "Not", so if A = 0, $\lnot$A = 1, same thing as B and $\lnot$B. Which brings me to that A and B have to be 0 or 1, otherwise $\lnot A \iff \lnot B$ with be false.
The symbol $\iff$ means if inputs are both true or false. So if A and b = 0 or 1, then A $\iff$ B is true otherwise false.

The $\lnot$ means "Not", so if A = 0, $\lnot$A = 1, same thing as B and $\lnot$B. Which brings me to that A and B have to be 0 or 1, otherwise $\lnot A \iff \lnot B$ with be false.