I have a structure $\mathscr A$ with domain of discourse the set of even natural numbers N and a language of two predicate symbols r, s which are interpreted in the following way: $$ <a,b,c> \in r^{\mathscr A} \iff a + b + 2 = c \\ <a,b,c> \in s^{\mathscr A} \iff ab=c $$
I have to define the singleton sets {0}, {2}, {4}, {6} and {$n \in N$: n is an even number which is not divided by 4 }
My solution is the following:
$$ \phi_0(x) \equiv \forall y \ s(x,y,x) \\ \phi_2(x) \equiv \exists y(\phi_0(y) \land r(y,y,x)) \\ \phi_4(x) \equiv \exists y \exists z(\phi_0(y)\land \phi_2(z) \land r(y,z,x)) \\ \phi_6(x) \equiv \exists y(\phi_2(y) \land r(y,y,x)) \\ \phi_{\sigma} \equiv \exists y(\phi_4(y) \land \neg \exists z \ s(z,x,y)) $$
Is that a proper solution? Should there be anything else added?
Thanks in advance!
Right idea, but a small mistake, since you have the product as 4, but 4 should be a divisor. So:
$$\phi_{\sigma}(x) \equiv \exists y(\phi_4(y) \land \neg \exists z \ s(z,y,x)) $$
And while your idea certainly works, maybe the following is a bit more intuitive just because it puts the emphasis on the negation:
$$\phi_{\sigma}(x) \equiv \neg \exists y \exists z (\phi_4(y) \land s(z,y,x)) $$
You can also exploit the fact that your domain is all even numbers:
$$\phi_{\sigma}(x) \equiv \neg \exists v \exists w \exists y \exists z (s(v,w,y) \land s(y,z,x))$$