Q.Simplify the symbolic form of the switching circuit $(p \wedge q) \vee (\sim p \wedge q)$ using the algebra of statements, and state the simplest equivalent circuit.

Factor out the common switch $q$, then use the complement and identity laws.

The circuit's symbolic form is $(p \wedge q) \vee (\sim p \wedge q)$.

Simplify. The common factor is $q$. By the distributive law (taking $q$ out): $(p \wedge q) \vee (\sim p \wedge q) \equiv (p \vee \sim p) \wedge q.$ By the complement law $p \vee \sim p \equiv \mathbf{t}$, and by the identity law $\mathbf{t} \wedge q \equiv q$: $(p \vee \sim p) \wedge q \equiv \mathbf{t} \wedge q \equiv q.$

So the circuit is equivalent to a single switch $q$ — current flows exactly when $q$ is closed, whatever the state of $p$. …

A common error is to factor incorrectly, e.g. taking p common instead of the genuinely shared q, or forgetting that p or ~p simplifies to t (always true) rather than to some p-dependent expre …