Case-I If ∆≡∇≡∧ (p∧q)→((p∧q)∧r) it can be false if r is false, so not a tautology Case-II If ∆≡∇≡V (p∨q)→((p∨q)∨r)≡ tautology then (p∨q)∨r≡(p∆r)∨q Case-III if ∆=∨,∇=∧ then (p∧q)→{(p∨q)∧r} Not a tautology ( Check p→T,q→T,r→F) Case-IV if ∆=∧,∇=∨ (p∧q)→{(p∧q)∨r} Not a tautology