Q.Prove De Morgan's law $\sim(p \vee q) \equiv \sim p \wedge \sim q$ using truth tables.

Tabulate $\sim(p \vee q)$ and $\sim p \wedge \sim q$ side by side and check the two columns agree.

Build the full table with all needed intermediate columns.

Compare. The column $\sim(p \vee q)$ is $F, F, F, T$. The column $\sim p \wedge \sim q$ is: row 1 $F \wedge F = F$; row 2 $F \wedge T = F$; row 3 $T \wedge F = F$; row 4 $T \wedge T = T$, i.e. $F, F, F, T$.

The columns match in every row, so $\sim(p \vee q) \equiv \sim p \wedge \sim q.$ …

The classic De Morgan error is to negate each part but keep the same connective, writing ~(p or q) as ~p or ~q. The connective must flip: negati …