Gate 2017 Set-1 first order logic

Question

Consider following first order logic

$\forall x ( \exists y R( x, y ) )$ is equivalent to

1) $\exists y ( \exists x R( x, y ) )$

2) $\exists y ( \forall x R( x, y ) )$

3) $\forall y ( \exists x R( x, y ) )$

4) $\neg \exists x ( \forall y \neg R(x,y) )$

Note: Not sure in the question was it equivalent to or which of these are implied by 

Comments

4) is straight forward. My doubt is can 1 is also be the ans?

Answer 1

let x=(4,6,8) y=(2,3,16) R(x,y) means x divide by y

 ∀x (∃y R(x, y))= means every x divide by some y which is hold true in our assumption (here 2 divide every x)

∃y (∃x R(x, y))= means some y divide by some x which is also hold true(here  16 divide by  4 or 8)

Answer 2

I think it was implied by.
1 and 4.

Comments

Yes me too. Sorry I just copied from online link.

PS: Apologies to everyone I dont want to make anyone nervous by posting wrong question.

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I think the ans will be 4th option one as 1st one says that "There exist y for all x" but question states "For all x there exist y"

For 4th option They have used negation. If we take outside negation inside then "There exist x" will change to "For All x" and "For All y" will change to "There exist y" which is same as given question......Hope this helps.....correct me if I'm wrong....peace...:-)

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@Anmol if the qn is equivalent to then you are right if it was implied by then both 1 and 4 should be correct

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Yup.....I got it....:-)

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@Anmol do you rem exactly was it equivalent or implied by?

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Actually I was on 2nd session......I just solved it because I want to learn.........:-)

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yes it seems to be 4

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Please explain why Option 1 is write?

Answer 3

straightforward 4 is equivalent
Also, option_2 implies the given statement., because if there is at least one y related to every x, conversely for all x there will obviously be some y.