Determining Statement Truth or Falsity? I have to determine if these are true or false, for "∀x∃y" I read it as "for all X there is a Y". The domain
consists of pairs x and y, where x is -1, 0, or 1 and y is -1, 0, or 1. My answers are on the right in italics, however I dont know if they are correct, could someone explain?
(a) ∀x∃y(x + y = 0)  true
(b) ∃x∀y(x + y = y)  true
(c) ∀x∃y(xy ≥ y)  true
(d) ∃x∀y(x ≤ y)  false
 A: It looks like all your answers are correct.
The qualifier $\forall x \exists y$ means "If you pick a value of $x$, I can always find a value of $y$ that satisfies the condition". So, for example, $\forall (x \in \mathbb{N}) \exists (y \in \mathbb{N}) \ (y = x + 1)$ means "For any natural number $x$, I can find another natural number $y$ that's one more than it.
So, if $x, y \in \{-1, 0, 1\}$, then we can actually check the truth of the statements explicitly by just checking "Is this true when $x = -1$? When $x = 0$? When $x = 1$?"
For example:
$\forall x \exists y (x + y = 0)$: When $x = -1$, we can pick $y = 1$. When $x = 0$, we can pick $y = 0$. When $x = 1$, we can pick $y = -1$. So this is true.
By comparison, $\exists x \forall y$ means "There is a single value of $x$ such that the statement is true for every possible $y$". So $\exists x \forall y (x + y = 0)$ is not true, because if $x = -1$ then $-1 + y$ is not always $0$, and the same is true for the other possible $x$ values. But thankfully, that's not the question you were asked to answer (I'm just using it for comparison).
$\exists x \forall y (x +  y = y)$ is true, because $x = 0$ works for all values of $y$.
