I need to do the negation of ∃y∀x(x ≥ y +1) using generalized de morgan's law to simplify the expression. This is what I have;

¬(∃y∀x(x ≥ y +1)) ≡ ∀y¬(∀x(x≥y +1)) ≡ ∀y∃x¬(x ≥ y +1) ≡ ∀y∃x(x < y +1)

I'm pretty sure I've done it all right except I'm not sure about the last statement in particular the < sign and the +1.

Okay that's my main problem that I would love to get some help on but if someone could also tell me if the following is correct.

For ∃y∀x(x ≥ y +1) I need to prove whether the statement is true or not if the domain of discourse is Z+ which I'm pretty sure I have the right answer for (this isn't what I want clarify) but the question is I need to state a pair of domains (one for x, one for y) for which the proposition is true. Would I be correct with the following?~

5 for x, 4 for y as 5 ≥ 4 + 1 ≡ 5 ≥ 5 is true.

Thanks for the help!