Let Δ , ∇ ∈ ∧ , ∨ be such that p ∇ q → p Δ q ∇ r is a tautology. Then p ∇ q ∆ r is logically equivalent to

Step-by-step solution

Case-I If Δ ≡ ∇ ≡ ∧ p ∧ q → p ∧ q ∧ r then it is false if r is false. So not a tautology Case-II If Δ ≡ ∇ ≡ ∨ p ∨ q → p ∨ q ∨ r ≡ tautology then p ∨ q ∨ r ≡ p Δ r ∨ q Case-III If Δ = ∨ , ∇ = ∧ then p ∧ q → p ∨ q ∧ r is not a tautology when p → T , q → T , r → F Case-IV If Δ = ∧ , ∇ = ∨ p &#