Concept:The expression is simplified using Boolean algebra laws such as distributive, complement, and identity laws.Explanation:Let E=((p∧q)∨(p∨∼q))∧(∼p∧∼q).Apply the distributive law to the first part:(p∧q)∨(p∨∼q)≡[p∨(p∨∼q)]∧[q∨(p∨∼q)].Using idempotent law, p∨p≡p, so p∨(p∨∼q)≡p∨∼q.Also, q∨(p∨∼q)≡p∨(q∨∼q)≡p∨T≡T.Therefore, the first bracket becomes (p∨∼q)∧T≡p∨∼q.So, E≡(p∨∼q)∧(∼p∧∼q).Distribute: (p∨∼q)∧(∼p∧∼q)≡(p∧∼p∧∼q)∨(∼q∧∼p∧∼q).Since p∧∼p≡F, the first term is F.The second term reduces to ∼p∧∼q.Thus, E≡F∨(∼p∧∼q)≡∼p∧∼q.Answer:The expression is equivalent to (∼p)∧(∼q), which is Option D.