1. State the problem: Verify De Morgan's Law $$\sim (p \wedge q) \equiv \sim p \vee \sim q$$ indicates that the negation of a conjunction is logically equivalent to the disjunction of the negations. 2. Define the truth values for $p$ and $q$, where each can be true ($T$) or false ($F$). 3. Evaluate the left side for each combination: - When $p=T$, $q=T$: $$p \wedge q = T \wedge T = T,$$ so $$\sim (p \wedge q) = \sim T = F.$$ - When $p=T$, $q=F$: $$p \wedge q = T \wedge F = F,$$ so $$\sim (p \wedge q) = \sim F = T.$$ - When $p=F$, $q=T$: $$p \wedge q = F \wedge T = F,$$ so $$\sim (p \wedge q) = \sim F = T.$$ - When $p=F$, $q=F$: $$p \wedge q = F \wedge F = F,$$ so $$\sim (p \wedge q) = \sim F = T.$$ 4. Evaluate the right side for each combination: - When $p=T$, $q=T$: $$\sim p = F, \sim q = F,$$ so $$\sim p \vee \sim q = F \vee F = F.$$ - When $p=T$, $q=F$: $$\sim p = F, \sim q = T,$$ so $$\sim p \vee \sim q = F \vee T = T.$$ - When $p=F$, $q=T$: $$\sim p = T, \sim q = F,$$ so $$\sim p \vee \sim q = T \vee F = T.$$ - When $p=F$, $q=F$: $$\sim p = T, \sim q = T,$$ so $$\sim p \vee \sim q = T \vee T = T.$$ 5. Compare both sides for every case: they match perfectly. 6. Conclusion: De Morgan's Law is verified as $$\sim (p \wedge q) \equiv \sim p \vee \sim q$$ holds true for all truth values of $p$ and $q$.