[Discrete maths] Predicate logic

Question

Which of the following is true about below predicate logic P?

A) P is satisfiable

B) P is tautology

C) P is contradiction

D) None

This expression in the end reduces to:-

~ ∀z { True }

Now this should mean => 

∃z { False}

So, how can it be contradiction as given answer is contradiction

Comments

Because there is no predicate inside to hold any value of Z therefore the complete predicate boils down to false only.

Answer 1

The questions says, 

This evaluates to there exists a z for which $\sim \left ( P(z) \rightarrow (\sim Q(z) \rightarrow P(z)) \right )$  is True.

But $\sim \left ( P(z) \rightarrow (\sim Q(z) \rightarrow P(z)) \right )$ is always False.

Therefore, saying that  $\exists z \sim \left ( P(z) \rightarrow (\sim Q(z) \rightarrow P(z)) \right )$, is False for all cases as there exists no such z. Therefore, this is a contradiction.