edited by
2,293 views
0 votes
0 votes

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 

edited by

3 Answers

2 votes
2 votes
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)
0 votes
0 votes
I think it was implied by.
1 and 4.
0 votes
0 votes
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.

Related questions

1 votes
1 votes
1 answer
2