This is an exercise problem in the textbook A Friendly Introduction to Mathematical Logic. The question is as follows:

Let $\mathfrak{N}$ be the usual structure of the language of number theory $\mathcal{L}_{NT}$. Let $\Sigma$ be the theory of $\mathfrak{N}$. Let $\mathcal{L}$ be a language given by $\mathcal{L}=\mathcal{L}_{NT} \cup \{ c \}$. Let $\Theta=\Sigma\cup \{ 0<c, S0<c, SS0<c,\ldots \}$. Let $\mathfrak{A}'$ be a model of $\Theta$. Now we define $\mathfrak{A}=\mathfrak{A}'\upharpoonright_{\mathcal{L}_{NT}}$. Prove that $\mathfrak{A}$ is elementarily equivalent to $\mathfrak{N}$.

My attempt:

In order to prove that $\mathfrak{A}$ is elementarily equivalent to $\mathfrak{N}$, we need to prove that theory of $\mathfrak{A}$ is $\Sigma$. Let's take $\phi\in \Sigma$. Then $\phi\in\Theta$. Thus, $\Theta \vdash \phi$ and by Soundness theorem, $\Theta\models\phi$. Since $\mathfrak{A}'$ is a model of $\Theta$, we have that $\mathfrak{A}'\models\phi$. Thus $\mathfrak{A}\models \phi$ as $\phi$ is a $\mathcal{L}_{NT}$ formula. Thus, $\Sigma$ is a subset of theory of $\mathfrak{N}$.

I am having trouble proving the other direction however. If I take a $\phi$ such that $\mathfrak{A}\models\phi$. Then I can say that $\mathfrak{A}'\models \phi$. But I am not really sure how to proceed after this.

Any hints will be appreciated.


HINT: $\Sigma$ is already "as big as possible" - if $\mathfrak{A}'\models\varphi$ for some $\varphi\not\in\Sigma$, there will be a "clash."

In more detail, suppose $\mathfrak{A}'\models\varphi$ but $\mathfrak{N}\not\models\varphi$. Can you think of a sentence related to $\varphi$ which $\mathfrak{N}$ does satisfy instead?

Its negation, $\neg\varphi$. Remember that by definition of the satisfaction relation we have $$\mathfrak{S}\not\models\eta\quad\iff\quad\mathfrak{S}\models\neg\eta$$ for every structure $\mathfrak{S}$ and every sentence $\eta$.

Now why would that be a problem?

If $\mathfrak{N}\models\neg\varphi$ then $\neg\varphi\in\Sigma$ so $\mathfrak{A}'\models\neg\varphi$. But this contradicts the assumption that $\mathfrak{A}'\models\varphi$.

Basically, what's going on here is that the theory of a structure is always a maximal consistent theory (with respect to the structure's language, that is), so there's no "room for improvement."

  • $\begingroup$ "The theory of a structure is always a maximal consistent theory" is the takeaway here. $\endgroup$ – ashK Mar 17 at 12:10

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