1. Construct the truth tables for the given propositions. **i.** $\sim p \lor q$ | $p$ | $q$ | $\sim p$ | $\sim p \lor q$ | |---|---|-------|------------| | T | T | F | T | | T | F | F | F | | F | T | T | T | | F | F | T | T | **ii.** $p \land \sim q$ | $p$ | $q$ | $\sim q$ | $p \land \sim q$ | |---|---|-------|--------------| | T | T | F | F | | T | F | T | T | | F | T | F | F | | F | F | T | F | 2. Show that $(p \Rightarrow q) \Rightarrow (\sim p \Rightarrow \sim q)$ is logically equivalent to $p \lor \sim q$. Rewrite implications: - $p \Rightarrow q \equiv \sim p \lor q$ - $\sim p \Rightarrow \sim q \equiv p \lor \sim q$ Expression: $$ (p \Rightarrow q) \Rightarrow (\sim p \Rightarrow \sim q) \equiv (\sim p \lor q) \Rightarrow (p \lor \sim q) $$ Rewrite the main implication: $$ \sim (\sim p \lor q) \lor (p \lor \sim q) \equiv (p \land \sim q) \lor (p \lor \sim q) $$ Simplify: $$ (p \land \sim q) \lor p \lor \sim q = p \lor \sim q $$ Thus, $$ (p \Rightarrow q) \Rightarrow (\sim p \Rightarrow \sim q) \equiv p \lor \sim q $$ 3. Construct truth table to verify: $$ \sim ((p \lor q) \land r) \equiv (\sim p \land \sim q) \lor \sim r $$ | $p$ | $q$ | $r$ | $p \lor q$ | $(p \lor q) \land r$ | $\sim ((p \lor q) \land r)$ | $\sim p$ | $\sim q$ | $\sim p \land \sim q$ | $\sim r$ | $(\sim p \land \sim q) \lor \sim r$ | |---|---|---|----------|-----------------|------------------------|-------|-------|----------------|-------|------------------------------| | T | T | T | T | T | F | F | F | F | F | F | | T | T | F | T | F | T | F | F | F | T | T | | T | F | T | T | T | F | F | T | F | F | F | | T | F | F | T | F | T | F | T | F | T | T | | F | T | T | T | T | F | T | F | F | F | F | | F | T | F | T | F | T | T | F | F | T | T | | F | F | T | F | F | T | T | T | T | F | T | | F | F | F | F | F | T | T | T | T | T | T | The columns $\sim ((p \lor q) \land r)$ and $(\sim p \land \sim q) \lor \sim r$ match, verifying the equivalence. 4. Determine whether the compound propositions are tautologies, contradictions, or contingent. I. $p \land \sim q \Rightarrow q \lor p$ Truth table indicates it is always true (tautology) because if $p \land \sim q$ is true, $p$ is true, so $q \lor p$ is true. II. $\sim (p \land r) \Leftrightarrow \sim (r \land p)$ This is equivalence of negation of conjunction symmetric in $p$ and $r$, always true (tautology). III. $p \Leftrightarrow (\sim p \land q)$ This expression is neither always true nor always false; it is true in some cases and false in others, so it is contingent. **Final answers:** - 1i. truth table shown - 1ii. truth table shown - 2. Logical equivalence verified - 3. Equivalence verified by truth table - 4. I and II are tautologies; III is contingent