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

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

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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.

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    look at here, please. http://math.stackexchange.com/questions/2109032/two-factorial-formulas-in-second-order-logic – 2017-01-23