This is an elementary question that I may blush about later, but for now: given that a recursively enumerable set is a set modeling a Σ01 sentence, and a recursive set is a recursively enumerable set S whose complement ℕ\S is also recursively enumerable. Fine. But then, letting x→ = the vector (x0, x1,...xn) S would be modeling a sentence of the form (*) ∃(x→)F(x→) where none of the xi's appear free in F. However, S is Δ01, so that it also models some statement of the form (**) ∀(x→) G(x→) I am not sure how one comes to the conclusion about Δ; I see that ℕ\S would model a sentence of the form ∃(x→) ~F(x→), (where ~ is negation), that is, ~∀(x→) F(x→) (I am not working in an intuitionist setting.) which doesn't quite make it. What silly mistake am I making, or what obvious fact am I overlooking? Thanks.