Made Easy mock test 1 Q:16 (2017-18)

Question

Which of the following is true for the predicate logic P ?

~ $\forall z[P(z) \rightarrow ($~$Q(z)\rightarrow P(z)) ]$

a.) P is satisfiable

b.) P is Tautology

c.) P is Contradiction

d.) None of these

Comments

Now check...

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But I have propagated the negation all the way till end in the expression

I have negated Q an P as well

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The expression in square brackets is true. It's a tautology.

But how to evaluate ~ ∀z [tautology] in logic terms?