Is there a structure with these properties?

Question

I'm looking for a structure which holds these properties:

For some domain, (eg. N,R, sets or whatever), there is a predicate P(x,y):

for all x from domain there is not y such that P(x,y)

AND either a) or b) is valid but not both:

a) there exists y such that for all x P(y,x)

OR

b) for all x there exists y such that P(y,x)

I've tried many P(x,y), for example P(x,y) => x mod y = 0 etc. but still I'm not able to find such structure.

Could you help me? Is there some?

EDIT: Repaired f to P - predicate, Predicate could be for example P(x,y)=> x mod y = 0

Answer 1

The "first statement" I interpret as $\forall x\exists y \neg P(x,y)$ (as if we interpret it as you have written $\forall x\neg\exists y P(x,y)$ then P needs to be the empty relation).

As your formulation is vague and weird I have two choices of answers for you:

*If I assume that $a)$ is $\exists y\forall x P(x,y)$ and $b)$ is $\forall x\exists y P(x,y)$. In this case let our structure have domain the natural numbers and say that $P(a,b)$ hold if ($a=0$ and $b=0$) or ($b=1$ and $a\neq 0$) In this case $a)$ is not satisfied but both $b)$ and the first statement is satisfied.

*On the other hand If we interpret your statement $a)$ as $\exists y \forall x P(y,x)$ and $b)$ as $\forall x \exists y P(y,x)$ then the negation of $a)$ will be equivalent with $\forall y\exists x \neg P(y,x)$ i.e. the first statement. Thus we can't have $a)$ to be true, hence $b)$ needs to be true. For this to be satisfied we may though just choose the opposite structure of what we previously did i.e. let P(a,b) hold if $a=0$ and $b=0$ or $b\neq 0$ and $a=1$

Answer 2

NO, there isn't.

Consider :

1) $\forall x \lnot \exists y P(x,y)$

2a) $\exists y \forall x P(y,x)$

2b) $\forall x \exists y P(y,x)$.

Case I : consider 2a) and let $a$ such that : $\forall x P(a,x)$. By $\forall$-E we have : $P(a,a)$.

But from 1) we get the equivalent : $\forall x \forall y \lnot P(x,y)$, and thus : $\lnot P(a,a)$. Contradiction !

Case II : consider 2b), and assume more than one element in the domain (the case with only one element is like I); we have : $P(b,a)$, but again from 1) we get $\lnot P(b,a)$. Contradiction !