Q.Obtain the simplest logical expression of the following and draw the corresponding switching circuit: $(\sim p \wedge q) \vee (\sim p \wedge \sim q) \vee (p \wedge \sim q)$

Start with $(\sim p \wedge q) \vee (\sim p \wedge \sim q) \vee (p \wedge \sim q)$.

Step 1 — factor $\sim p$ out of the first two terms (distributive law): $\sim p \wedge (q \vee \sim q) \vee (p \wedge \sim q)$.

Step 2 — complement law, $q \vee \sim q \equiv T$, then identity law, $\sim p \wedge T \equiv \sim p$: expression becomes $\sim p \vee (p \wedge \sim q)$.

Step 3 — distribute: $\sim p \vee (p \wedge \sim q) \equiv (\sim p \vee p) \wedge (\sim p \vee \sim q) \equiv T \wedge (\sim p \vee \sim q) \equiv \sim p \vee \sim q$. …

Factor a common literal, apply complement + identity laws, then …

Stopping after step 2 at $\sim p \vee (p \wedge \sim q)$ without applying the final absorption/distributive step, or wrongl …