Q.Construct the truth table for: $(\sim p \vee \sim q) \leftrightarrow [\sim(p \wedge q)]$

Build the table over all combinations of $p,q$:

$p=T,q=T$: $\sim p\vee\sim q = F\vee F=F$; $p\wedge q=T$, $\sim(p\wedge q)=F$; biconditional $F\leftrightarrow F=T$.

$p=T,q=F$: $\sim p\vee\sim q=F\vee T=T$; $p\wedge q=F$, $\sim(p\wedge q)=T$; $T\leftrightarrow T=T$.

$p=F,q=T$: $\sim p\vee\sim q=T\vee F=T$; $p\wedge q=F$, $\sim(p\wedge q)=T$; $T\leftrightarrow T=T$.

$p=F,q=F$: $\sim p\vee\sim q=T\vee T=T$; $p\wedge q=F$, $\sim(p\wedge q)=T$; $T\leftrightarrow T=T$. …

Compute both sides of the biconditional independently, then com …

Assuming a biconditional needs a mix of T/F rows to be 'interesting' — here it is a tautology prec …