Then (p∆∼q)∨(∼p∆q) becomes (p∨∼q)∨(∼p∨q) which is always true, so x becomes a tautology. Case-II When ∆ is same as ∧ Then (p∧q)⇒(p∧∼q)∨(∼p∧q) If p∧q is T, then (p∧∼q)∨(∼p∧q) is F so x cannot be a tautology. Case-III When ∆ is same as ⇒Then (p⇒∼q)∨(∼p⇒q) is same at (∼p∨∼q)∨(p∨q), which is always true, so x becomes a tautology. Case-IV When ∆ is same as ⇔ Then (p⇔q)⇒(p⇔∼q)∨(∼p⇔q) p⇔q is true when p and q have same truth values, then p⇔∼q and ∼p⇔q both are false. Hence x cannot be a tautology.