Let p,q,r be three statements, then (p→(q→r))↔((p∧q)→r), is a
tautology
contradiction
fallacy
none of these
p→(q→r)≡∼p∨(q→r)
≡−p∨(−q∨r)≡[(~p)∨(~q)]∨r≡∼(p∧q)∨r≡p∧q→r
∴ given statement is a tautology.