P↔Q = P Exnor Q
So ~ ( P Exnor Q) = ~ P Exnor Q = P Exnor ~Q = P Exor Q
Yes both are true.
Both option(1)and (2) is the correct choice.
....
$\sim$(P $\rightarrow$ Q) = $\sim$( (P $\rightarrow$ Q) ^ (Q $\rightarrow$ P) ) = $\sim$ ( ($\sim$P v Q) ^ ( $\sim$Q v P) ) = $\sim$($\sim$P v Q) v $\sim$($\sim$Q v P) = (P ^ $\sim$Q) v (Q v $\sim$P) = P$\bigoplus$Q Using distributive law. (a + bc) = (a + b).(b + c) = ( (P ^ $\sim$Q) v Q) ^ ( (P ^ $\sim$Q) v $\sim$P) = ( ( P v Q)^($\sim$Q v Q)) ^ ( ( $\sim$P v P) ^ ($\sim$Q v $\sim$P) ) = ( ( P v Q) ^ ( $\sim$Q v $\sim$ P) Now here we have two choices. 1. ($\sim$P $\rightarrow$ Q) ^ ( Q $\rightarrow$ $\sim$P) = $\sim$P $\leftrightarrow$ Q 2. ($\sim$Q $\rightarrow$ P) ^ (P $\rightarrow$ $\sim$ Q) = P $\leftrightarrow$ $\sim$Q.
So both 1 and 2 option is correct.