Let $\sigma$ be a sentence (in the language of $\mathfrak{N}$ ) in the prenex normal form (Section 2.6). We use induction on the number of quantifiers in $\sigma$ to show that $T\vDash\sigma$ iff $\sigma$ is true in $\mathfrak{A}$ . We already know by the preceding exercise that for every quantifier-free sentence $\sigma$ , if $\sigma$ is true in $\mathfrak{A}$ , then $A_{E}\vdash\sigma$ , hence, $T\vDash\sigma$ , and if $\sigma$ is false in $\mathfrak{A}$ , then $A_{E}\vdash\neg\sigma$ , hence, $T\vDash\neg\sigma$ . If $\sigma=\forall x\phi$ is true in $\mathfrak{N}$ , where $x$ is the only free variable in $\phi$ , then for all $n\in\mathbb{N}$ , $\vDash_{\mathfrak{A}}\phi[s(x|n)]$ , and, by hypothesis, $T\vDash\phi_{\mathbb{S}^{n}0}^{x}$ , but then, since $T$ is $\omega$ -complete, $T\vDash\forall x\phi=\sigma$ . If $\sigma=\exists x\phi$ is true in $\mathfrak{N}$ , then for some $n\in\mathbb{N}$ , $\vDash_{\mathfrak{A}}\phi[s(x|n)]$ , and, by hypothesis, $T\vDash\phi_{\mathbb{S}^{n}0}^{x}$ , but then $T\vDash\exists x\phi=\sigma$ . Now, if $\sigma=\forall x\phi$ or $\sigma=\exists x\phi$ is false in $\mathfrak{A}$ , then $\neg\sigma$ (which is logically equivalent to $\exists x\neg\phi$ or $\forall x\neg\phi$ , respectively) is true in $\mathfrak{A}$ , and, hence, $T\vDash\neg\sigma$ . Therefore, by induction, we conclude that $T\vDash\sigma$ ($\sigma\in T$ ) iff $\sigma$ is true in $\mathfrak{A}$ ($\sigma\in\mbox{Th}\mathfrak{N}$ ) (here, we use the fact the $T$ is consistent).

Note. $\mbox{Cn}A_{E}$ is not $\omega$ -complete. For example, if $\phi=(y\neq0\rightarrow\exists xy=\mathbb{S}x)$ , then $A_{E}\vdash\phi_{\mathbb{S}^{n}0}^{y}$ for every $n\in\mathbb{N}$ , however, $A_{E}\nvdash\forall y(y\neq0\rightarrow\exists xy=\mathbb{S}x)$ , which is axiom S3 of $A_{S}$ .