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There is the common construction of a non-standard model of arithmetic by adding a constant symbol c to the signature and adding $\{n<c:n\in \mathbb{N}\}$ 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?

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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 $c$.

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  • $\begingroup$ So elementary equivalence (and isomorphy) of models are always relative to a signature? $\endgroup$ – GottlobtFrege Jan 11 '19 at 0:03
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    $\begingroup$ Yes, though the signature usually is not stated explicitly. $\endgroup$ – Eric Wofsey Jan 11 '19 at 0:04
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    $\begingroup$ Of course we can still talk about the order-type, since $<$ is still part of the signature. $\endgroup$ – Eric Wofsey Jan 11 '19 at 0:09
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    $\begingroup$ @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. $\endgroup$ – Noah Schweber Jan 11 '19 at 0:14
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    $\begingroup$ @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). $\endgroup$ – spaceisdarkgreen Jan 11 '19 at 2:03

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