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

Contradiction?

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Yes given ans is contradiction but my doubt is if I am propagating the negation inside

it would become

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

so further solving this it comes

$\exists z$ (True)

what is wrong with this approach ?

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∀z [A(z)]  -> For all z, A[z] is true

~ would be ∃z for which A[z] will be false.

I thought in statements. Taking negation in terms of statement would be

for all z A[z] is false.

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if everyone is true its negation would be atleast one is false

like in a village everyone says truth only

so for negating this statement atleast one person saying lie would suffice

 

hope you get it

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~∀z[P(z)→(~Q(z)→P(z))]

you are solving problem in your ways (by creating new thing). when you propagate negation then why doing partiality with inside expression.

~ ∀z[  some thing]

what you are solving is 

~ ∀z[P(z)]→(~Q(z)→P(z))

Hope you got it.

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sorry, I didn't get you. your expression seems different from what I have posted .

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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?