Consider the structure $A=<\mathbb{N},<;\cdot,1>$. Find the following interpretations of formulas (for an arbitrary valuation v):
(c) $[\![ \forall x_1\exists x_1 P(x_1,x_2) ]\!]$
(d) $[\![ \exists x_3\forall x_4 P(x_1,x_2) ]\!]$
Solution:
(c) $$[\![ \forall x_1\exists x_1 P(x_1,x_2) ]\!]$$ $$=\begin{cases} 1 \text{ if for all } a\in |A|\text{ there is some } b\in |A|\text{ } [\![ P(x_1,x_2) ]\!]^{[x_1\to a][x_1\to b]}\\ 0 \text{ otherwise}\\ \end{cases}$$ $$[\![ P(x_1,x_2) ]\!]^{[x_1\to a][x_1\to b]}$$ $$=P^A([\![ x_1 ]\!]^{[x_1\to a][x_1\to b]},[\![ x_2 ]\!]^{[x_1\to a][x_1\to b]})$$ $$=P^A(b,v(x_2))$$ $$=b<v(x_2).$$
When I have arrived at this step I don't really know if I should say it's true or false. in (a) and (b) it was obvious since I had $\forall x_1\exists x_2(...)$ but i haven't said anything about the existence of $x_2$ in this particular case, what is $v(x_2)$ in this case (which feels kinda weird to me)? Is it just an arbitrary valuation of the natural numbers, that is, that I can just say this is true? Same goes for (d) then I get $v(x_1)<v(x_2)$ is this also true? It was easier in (a) since I had "for all $a\in |A|$ there is some $b\in |A|$ such that $a<b$. If I haven't said that there exists an $x_2$ with this property nor have I said for all $x_2$ how can I then draw a conclusion?