1. The problem asks to verify logical equivalences and simplify logical expressions. (a) Show that $ (p \lor q) \to p $ is logically equivalent to $ (\sim p \land \sim q) \lor p $ using a truth table. Create columns for $p$, $q$, $p \lor q$, $\sim p$, $\sim q$, $ (p \lor q) \to p $, and $ (\sim p \land \sim q) \lor p $. Calculate each: - $p \lor q$ is true if either $p$ or $q$ is true. - $\sim p$, $\sim q$ are negations. - $ (p \lor q) \to p $ is false only when $p \lor q$ is true but $p$ is false. - $ (\sim p \land \sim q) \lor p $ is true if either both $p$ and $q$ are false or $p$ is true. The truth table rows match for both expressions, showing equivalence. 2. (b) Draw the simplest electric network for $ q \lor (p \land \sim q) \lor (r \land \sim p) $. This represents a logical OR of these terms; - $q$ alone - $p$ AND NOT $q$ - $r$ AND NOT $p$ This can be modeled using switches: - One switch for $q$ in parallel. - One series connection for $p$ and NOT $q$ in parallel. - One series connection for $r$ and NOT $p$ in parallel. 3. (c) Simplify $ \sim (p \lor q) \lor (\sim p \land q) $. Use De Morgan's law: $\sim (p \lor q) = \sim p \land \sim q$. So the expression becomes: $$ (\sim p \land \sim q) \lor (\sim p \land q) $$ Factor out $\sim p$: $$ \sim p \land (\sim q \lor q) $$ Since $(\sim q \lor q)$ is a tautology (always true), expression simplifies to: $$ \sim p $$ 4. (d) Test the validity of the argument: - Premise 1: If remedial classes (R), then standard IV pupils understand lessons well (U): $R \to U$ - Premise 2: If $U$, then no failure (F'): $U \to F'$ - Premise 3: Failure exists ($F$). - Conclusion: No remedial classes ($\sim R$). From premises 1 and 2, $R \to U$ and $U \to F'$ implies $R \to F'$ (transitivity). Given $F$ (failure), not $F'$, contrapositive of $R \to F'$ is $F \to \sim R$. Since $F$ is true, $\sim R$ follows, so the argument is valid.