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

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

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

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

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

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

Build the table row-by-row through $\sim p,\sim q$, the inner AND, its nega …

Misapplying De Morgan's law with the wrong connective (writing $\sim(\sim p\wedge\sim q)$ as …