Question

The Boolean expression \((p \land q) \Rightarrow ((r \land q) \land p)\) is equivalent to :

• A. \((q \land r) \Rightarrow (p \land q)\)
• B. \((p \land q) \Rightarrow (r \land q)\)
• C. \((p \land q) \Rightarrow (r \lor q)\)
• D. \((p \land r) \Rightarrow (p \land q)\)

Hint:

When simplifying complex Boolean expressions, consistently applying the basic laws (De Morgan's, Distributive, Associative) is key. Converting implications (\(\Rightarrow\)) to disjunctions (\(\lor\)) using \(A \Rightarrow B \equiv \neg A \lor B\) is often the most effective first step.

Answer 1

The Correct Option is B

Step 1: Convert the implication to its equivalent disjunctive form.
We use the logical equivalence \(A \Rightarrow B \equiv \neg A \lor B\). Let the given expression be E. \[ E \equiv \neg(p \land q) \lor ((r \land q) \land p) \] Step 2: Apply logical laws to simplify the expression.
- Apply De Morgan's Law to the first part: \(\neg(p \land q) \equiv \neg p \lor \neg q\). - Apply Commutative and Associative Laws to the second part: \((r \land q) \land p \equiv p \land q \land r\). The expression becomes: \[ E \equiv (\neg p \lor \neg q) \lor (p \land q \land r) \] Step 3: Apply the Distributive Law.
We use the law \(A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)\). Let \(A = (\neg p \lor \neg q)\) and \(B \land C = (p \land q) \land r\). \[ E \equiv [(\neg p \lor \neg q) \lor (p \land q)] \land [(\neg p \lor \neg q) \lor r] \] The first part \([(\neg p \lor \neg q) \lor (p \land q)]\) is equivalent to \(\neg(p \land q) \lor (p \land q)\), which is a tautology (T). So the expression simplifies to: \[ E \equiv T \land ((\neg p \lor \neg q) \lor r) \equiv \neg p \lor \neg q \lor r \] Step 4: Analyze the options to find an equivalent expression.
Let's simplify option (B): \((p \land q) \Rightarrow (r \land q)\). \[ (p \land q) \Rightarrow (r \land q) \equiv \neg(p \land q) \lor (r \land q) \] \[ \equiv (\neg p \lor \neg q) \lor (r \land q) \] Apply the Distributive Law: \[ \equiv (\neg p \lor \neg q \lor r) \land (\neg p \lor \neg q \lor q) \] The second part \((\neg p \lor \neg q \lor q)\) is equivalent to \((\neg p \lor T)\), which is a tautology (T). So, the expression simplifies to: \[ \equiv (\neg p \lor \neg q \lor r) \land T \equiv \neg p \lor \neg q \lor r \] This matches our simplified form of the original expression. Therefore, the expressions are equivalent.