Solution

The correct answer is A, B & C Only.

Key Points

• A. [¬p ∧ (p ∨ q)] → q
  • Simplify antecedent: ¬p ∧ (p ∨ q) = (¬p∧p) ∨ (¬p∧q)
  • Using contradiction law: ¬p∧p = 0
  • So antecedent = ¬p∧q
  • Expression becomes: (¬p∧q) → q
  • Implication rule: A→B ≡ ¬A ∨ B
  • = ¬(¬p∧q) ∨ q = (p ∨ ¬q) ∨ q
  • = p ∨ (¬q ∨ q) = p ∨ 1 = 1
  • Hence A is tautology.
• B. ¬p → (p → q)
  • p→q ≡ ¬p ∨ q
  • Expression becomes: ¬p → (¬p ∨ q)
  • = ¬(¬p) ∨ (¬p ∨ q) = p ∨ ¬p ∨ q
  • = 1 ∨ q = 1
  • Hence B is tautology.
• C. [p ∧ (p → q)] → q
  • p→q ≡ ¬p ∨ q
  • p ∧ (¬p ∨ q) = (p∧¬p) ∨ (p∧q)
  • = 0 ∨ (p∧q) = p∧q
  • Expression becomes: (p∧q) → q
  • = ¬(p∧q) ∨ q = (¬p ∨ ¬q) ∨ q
  • = ¬p ∨ (¬q ∨ q) = ¬p ∨ 1 = 1
  • Hence C is tautology.
• D. ¬(p → q) → q
  • p→q ≡ ¬p ∨ q
  • ¬(p→q) = ¬(¬p ∨ q) = p ∧ ¬q
  • Expression becomes: (p∧¬q) → q
  • = ¬(p∧¬q) ∨ q = (¬p ∨ q) ∨ q
  • = ¬p ∨ q
  • This is NOT always true (p=1, q=0 ⇒ false).
  • Hence D is not tautology.

Additional Information

• Conclusion:
  • A, B and C evaluate to 1 for all truth values.
  • D becomes false for p=1, q=0.
  • Therefore correct option is A, B & C only.