This is an exercise in the book "A Friendly Introduction to Mathematical Logic" (Authors: Christopher Leary and Lars Kristiansen).
Let $N$ be the peano axioms without the induction axiom, i.e., there are 11 axioms:
- $(\forall x)\neg Sx=0.$
- $(\forall x)(\forall y)\left[Sx=Sy\longrightarrow x=y\right].$
- $(\forall x)x+0=x.$
- $(\forall x)(\forall y)x + Sx = S(x+y).$
- $(\forall x)x\cdot0=0.$
- $(\forall x)(\forall y)x\cdot Sy=(x\cdot y)+x.$
- $(\forall x)xE0=S0.$
- $(\forall x)(\forall y)xE(Sy)=(xEy)\cdot x.$
- $(\forall x)\neg x<0.$
- $(\forall x)(\forall y)\left[x<Sy\longleftrightarrow \left(x<y\vee x=y\right)\right].$
- $(\forall x)(\forall y)\left[\left(x<y\right)\vee\left(x=y\right)\vee\left(y<x\right)\right].$
The exercise asks for a structure $\mathfrak{M}$ such that $\mathfrak{M}\models N$ and $\mathfrak{M}\models\neg(\forall x)(\forall y)[x+y=y+x]$
I do not know whether it is a trivial question; I hope it is not. I have tried to construct this model, but I could not. I see the universe as matrices or functions from $\mathbb N$ to $\mathbb N$ and define $+$ as composition, but I could not define $S$ such that the axioms 1 and 4 are both satisfied.
I also thought that we could prove the commutativity for adding two numbers like $SS\dots S0$ in a finite number of steps (it is just intuition), so I added an element that does not have a predecessor. But I could not define the functions as wanted in this case either. My other attempts are also failed.