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

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

Step 1 — the first term $p \wedge q \wedge \sim p = (p \wedge \sim p) \wedge q \equiv F \wedge q \equiv F$ by the complement law, so it drops out.

Remaining: $(\sim p \wedge q \wedge r) \vee (p \wedge \sim q \wedge r) \vee (p \wedge q \wedge r)$.

Step 2 — every remaining term contains $r$, so factor it out: $r \wedge [(\sim p \wedge q) \vee (p \wedge \sim q) \vee (p \wedge q)]$.

Step 3 — inside the bracket, combine the last two terms: $(p \wedge \sim q) \vee (p \wedge q) \equiv p \wedge (\sim q \vee q) \equiv p \wedge T \equiv p$ (distributive, complement, identity laws).

So the bracket becomes $(\sim p \wedge q) \vee p$. By absorption ($A \vee (\sim A \wedge X) \equiv A \vee X$ with $A=p$, $X=q$): $(\sim p \wedge q) \vee p \equiv p \vee q$. …

Complement law removes the self-contradictory term; factor the common literal $r$; distributive + …

Forgetting to drop the first (self-contradictory) term before simplifying, or factoring $r$ incorrectly by also …