Answer C&D
Let Two Cases for P
1st Case P=True
S : (((q⇒r))⇔((q)⇒r))∧∼p∧∼q∧∼r As (True⇒Anything is Anything) and (True AND Anything is anything)
S : ((q⇒r)⇔(q⇒r))∧∼p∧∼q∧∼r ((q⇒r)⇔(q⇒r)) is always True As (P⇔P) is always True
S : T∧∼p∧∼q∧∼r
S : ∼p∧∼q∧∼r Option D
and p is True so ∼p is False. Then S is a Contradiction
2nd Case P=False
S : ((p⇒(q⇒r))⇔((p∧q)⇒r))∧∼p∧∼q∧∼r
False⇒Anything is True and False∧Anything is False
S : (T⇔(F⇒r))∧∼p∧∼q∧∼r
S : (T⇔T)∧∼p∧∼q∧∼r
S : T∧∼p∧∼q∧∼r
S : ∼p∧∼q∧∼r Option D
And P is False so ∼p is True. Then S : ∼q∧∼r which is a contingency Option C