Past exam of the model theory course of the University of Oxford / 2011 / 1 / b / i

Yes. In the additive first-order structure, the quantifier-free formulaholds exactly at zero, because cancellation in the additive group gives $x=0$. This first-order formula uses only the permitted function symbol and equality; it does not require a constant symbol for zero.

No. The map $s(x)=-x$ is an automorphism of a first-order structure for the additive integers, but it sends one to minus one. By automorphism invariance of parameter-free definability, every parameter-free definable singleton is fixed by every structure automorphism. The singleton containing one is not fixed, so it is not definable.

No. Definability of a function means definability of its graph. The same additive structure automorphism $s(x)=-x$ would have to preserve the graph of multiplication. It sends the graph point $(1,1,1)$ to $(-1,-1,-1)$, which is not a multiplication-graph point because $(-1)(-1)=1\ne -1$. This contradicts automorphism invariance of parameter-free definability.

No. Interpreting the printed ${\mathbb{Z}}^{⩾}$ as the nonnegative integers, negation is an additive structure automorphism carrying the positive integer one outside that set. The set is therefore not invariant under structure automorphisms and cannot have parameter-free definability. The same argument would apply if the notation were taken to mean the strictly positive integers.