A question from Enderton's Mathematical Introduction to logic

Question

Why we can't deduce S3 from AE? This is from Enderton's Mathematical Introduction to logic page 203 enter image description here

Answer 1

To show that $A_E$ cannot prove $S3$, it suffices to give a model of $A_E$ in which $S3$ is false.

The model will have domain $\mathbb{N}\times \{0,1\}$. The idea is that we have two copies of $\mathbb{N}$: $\{(n,0)\mid n\in \mathbb{N}\}$ and $\{(n,1)\mid n\in \mathbb{N}\}$. The elements of the form $(n,1)$ are all greater than the elements of the form $(m,0)$. The operations are the natural ones on each copy, but the second copy is "absorbing" in the sense that if an operation is applied to a pair of elements, one from each copy, the output is in the second copy (unless the second input is $0$).

Here's the formal definition:

• $\mathbf{0} = (0,0)$.
• $\mathbf{1} = (1,0)$.
• $(n,i) < (m,j)$ if and only if (a) $i<j$ or (b) $i = j$ and $n<m$.
• $S(n,i) = (n+1,i)$.
• $(n,i) + (m,j) = (n+m,\max(i,j))$.
• $(n,i) \cdot (m,j) = (nm,\max(i,j))$, unless $(m,j) = (0,0)$, in which case $(n,i) \cdot (0,0) = (0,0)$.
• $(n,i) \mathbf{E} (m,j) = (n^m,\max(i,j))$, unless $(m,j) = (0,0)$, in which case $(n,i) \mathbf{E} (0,0) = (1,0)$.

Now you can check that the axioms hold. But $S3$ does not hold, since $(0,1)\neq \mathbf{0}$, but there is no $(n,i)$ such that $\mathbf{S}(n,i) = (0,1)$.