If p and q are logical statements, then p ∧ q → p → q is equivalent to

Step-by-step solution

∵ s → t ≡ ~ s ∨ t ∴ p ∧ q → p → q ≡ ∼ p ∨ ∼ q ∨ ~ p ∨ q ≡ ~ p ∨ q ∨ ~ q ≡ ∼ p ∨ t ≡ t Also, p → p ∨ q ≡ ~ p ∨ p ∨ q ≡ t