Question

The expression $[(p \wedge \sim q) \vee q] \vee (\sim p \wedge q)$ is equivalent to

• A. $p \vee q$
• B. $p \wedge q$
• C. $p \rightarrow q$
• D. $p \leftrightarrow q$

Hint:

When resolving logical equivalences in an exam, constructing a quick truth table or substituting truth values can save time. For instance, if you set $p = \text{True}$ and $q = \text{False}$, the expression evaluates to $[(\text{True} \wedge \text{True}) \vee \text{False}] \vee (\text{False} \wedge \text{False}) = \text{True}$. Checking the options, $p \vee q$ is $\text{True}$, which immediately eliminates options (B) and (C).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Simplify the logical expression [(p ∧ ∼q) ∨ q] ∨ (∼p ∧ q).

Step 2: Key Formula or Approach:
Use Distributive Law: (A ∧ B) ∨ C ≡ (A ∨ C) ∧ (B ∨ C). Complement Law: A ∨ ∼A ≡ True. Identity Law: A ∧ True ≡ A.

Step 3: Detailed Explanation:
Inner bracket: [(p ∧ ∼q) ∨ q] = (p ∨ q) ∧ (∼q ∨ q) = (p ∨ q) ∧ True = p ∨ q. Original becomes: (p ∨ q) ∨ (∼p ∧ q) = [(p ∨ q) ∨ ∼p] ∧ [(p ∨ q) ∨ q] = [(p ∨ ∼p) ∨ q] ∧ [p ∨ q] = [True ∨ q] ∧ (p ∨ q) = True ∧ (p ∨ q) = p ∨ q.

Step 4: Final Answer:
The expression simplifies to p ∨ q, option (A).