# Elementary equivalence of standard and non-standard model of arithmetic

There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding $$\{n to the theory PA. Now by adding all sentences $$\sigma$$ st. $$\mathbb{N}\models \sigma$$ we get a theory $$T^*$$, which is consistent by compactness theorem. My textbook says now that any model of $$T^*$$ is a model of PA which is elementary equivalent to $$\mathbb{N}$$.

However, I found definitions of elementary equivalence only for models of the same language. The language of the model constructed above differs from the standard language $$\mathcal{L}_{PA}$$ by c.

Where am I going wrong?

Any model of $$T^*$$ can also be considered as a model of PA by just forgetting about $$c$$. In other words, you consider it as a structure with only the arithmetic operations and not the constant $$c$$.
• Of course we can still talk about the order-type, since $<$ is still part of the signature. – Eric Wofsey Jan 11 '19 at 0:09
• @GottlobtFrege We don't forget the element named by $c$, it's still part of the reduct, we just forget the special name for it (= corresponding constant symbol "$c$"). So we're not losing any elements - the "weird shape" of the nonstandard model (that is, its ordertype, which is different from that of $\mathbb{N}$) is still there. – Noah Schweber Jan 11 '19 at 0:14
• @GottlobtFrege Yes, it is common for two inequivalent structures to have equivalent reducts. This is an example: if we extend the signature of $(\mathbb N,+,\cdot,0,1)$ to include $c,$ and interpret $c$ as $27,$ then $\mathbb N$ is not e.e. to any of the models of $T^*,$ even though their reducts to $(+,\cdot,0,1)$ are. Also, take any two non-e.e. models of PA. Their reducts to $(+,0,1)$ will be elementarily equivalent (since Presburger arithmetic is complete and PA is an extension). – spaceisdarkgreen Jan 11 '19 at 2:03