I have $\mathcal A(\mathcal U, \mathcal I)$ with $U_{\mathcal A} = \Bbb N$ and $I_{\mathcal A}(P) = \{(m,n) \mid m,n \in \Bbb R, m=\alpha n, \alpha \in \Bbb N \setminus\{0,1\}\}$
I need to show if that structure is a model for $F = \forall z \exists y \forall x (P(x,y) \lor P(z,y)\lor P(x,z) \lor \lnot P(z,x))$
I started writing down the meaning of said formula and got
For all $z$ there exists an $y$ for which all $x$, either $(\alpha y = x)$ or $(\alpha y= z)$ or $(\alpha z = x)$ or $(\alpha x \neq z)$
- Is that formulation correct?
- For the structure to be a model, do I need to find a single assignment for $x,y,z$ that satisfies the formula?
How do the quantifiers act on the assignment of the variables?
For the formula to be satisfied, since there are disjunctive connectors, only one of them needs to be satisfied right? Especially in respect to the variables and quantifiers
As always, thank you very much!
Yes. Your sentence is a straightforward translation.... though as such it does not bring any further insight into what exactly the statement is claiming ... I would provide some intuitive meaning for the $P$ relation ... how about 'multiple of'? You can also try to rewrite the formula into somethng more easily understood ... how about $P(z,x) \rightarrow (P(x,y) \lor P(z,y) \lor P(x,z))$?
If the statement was $\exists z \exists y \exists x...$, then yes, ... but the statement is $\forall z \exists y \forall x...$, so to satisfy the statement you need to consider all possible instances of $x$ and $z$.
... not exactly sure what you're asking. A universal means that it considers all possible assignments, and an existential claims something for some assignment.
That's right. Of the disjunction with the $4$ disjunct you need at least to be true ... so even if you have only one satisfied, you're good.