*,⊙∈{∧,∨} Now for (p*q)⊙(p⊙∼q) is tautology (A) (∨,∧):(p∨q)∧(p∧∼q) not a tautology (B) (∨,∨):(p∨q)∨(p∨∼q) =P∨T is tautology (C) (∧,∧):(p∧q)∧(p∧∼q) =(p∧p)∧(q∧∼q)=p∧F not a tautology (Fallasy) (D) (∧,∨):(p∧q)∨(p∨∼q) not a tautology