Equivalence of two formulas in FO.

Question

$$\phi(x,y) = \exists u \exists v u < x \wedge v < y \wedge f(u,v)$$ $$\phi(x,y) = \forall u \forall v (u < x \wedge v < y) \implies f(u,v)$$

Are above formulas equivalent? Why, intuitevly? Why, formally?

Answer 1

Let be $\Bbb N$ the universe, $<$ the usual order, $f(u,v)$ always true. The implication $(u < x \wedge v < y) \implies f(u,v)$ will be always true. But $$\phi(0,0) = \exists u \exists v u < 0 \wedge v < 0 \wedge f(u,v)$$ is false.