Logical Equivalence 4Cb4Bf

1. **State the problem:** Show that the compound propositions (i) $p \leftrightarrow q$ and $\sim(p \lor q) \lor (p \land q)$ (ii) $\sim(t \lor s) \lor (\sim t \land s)$ and $\sim t$ are logically equivalent without using truth tables. 2. **Recall definitions and equivalences:** - The biconditional $p \leftrightarrow q$ is logically equivalent to $(p \land q) \lor (\sim p \land \sim q)$. - De Morgan's laws: $\sim(p \lor q) \equiv \sim p \land \sim q$. - Distributive and associative laws help simplify expressions. --- ### Part (i): Show $p \leftrightarrow q$ is equivalent to $\sim(p \lor q) \lor (p \land q)$ 3. **Rewrite $p \leftrightarrow q$:** $$p \leftrightarrow q \equiv (p \land q) \lor (\sim p \land \sim q)$$ 4. **Rewrite $\sim(p \lor q) \lor (p \land q)$ using De Morgan's law:** $$\sim(p \lor q) \lor (p \land q) \equiv (\sim p \land \sim q) \lor (p \land q)$$ 5. **Compare both expressions:** Both are $(p \land q) \lor (\sim p \land \sim q)$, so they are logically equivalent. --- ### Part (ii): Show $\sim(t \lor s) \lor (\sim t \land s)$ is equivalent to $\sim t$ 6. **Rewrite $\sim(t \lor s)$ using De Morgan's law:** $$\sim(t \lor s) \equiv \sim t \land \sim s$$ 7. **Substitute into the expression:** $$\sim(t \lor s) \lor (\sim t \land s) \equiv (\sim t \land \sim s) \lor (\sim t \land s)$$ 8. **Factor out $\sim t$:** $$= \sim t \land (\sim s \lor s)$$ 9. **Since $\sim s \lor s$ is a tautology (always true):** $$\sim t \land \text{True} = \sim t$$ 10. **Therefore:** $$\sim(t \lor s) \lor (\sim t \land s) \equiv \sim t$$ --- **Final answers:** - (i) $p \leftrightarrow q$ is logically equivalent to $\sim(p \lor q) \lor (p \land q)$. - (ii) $\sim(t \lor s) \lor (\sim t \land s)$ is logically equivalent to $\sim t$. These equivalences are shown using logical identities and laws without truth tables.