If p and q are two statements, then ∼(p∧q)∨∼(q⇔p) is
tautology
contradiction
neither tautology nor contradiction
none of these
S=∼(p∧q)∨∼(q⇔p) Pq∼(p∧q)∼(q⇔p)STTFFFTFTTTFTTTTFFTFT
Statement (p∧∼q)∧(∼p∨q) is