Show that the predicate $y=x+1$ is not definable in the interpretation $(\mathbb Z,=,f)$ where $f$ is a unary function $x \mapsto x+2$

Question

I tried to build an automorphism that preserves the equality relation and the function $f$, such that if $l$ is that automorphism then $l(f(x)) = f(l(x))$ and such that if we define $P(x,y)$ to be the predicate $y=x+1$ then $P(x,y)$ won't equal to $P(l(x),(l(y))$.

I couldn't really find one. I was trying to argue that if I define $l(x):= x+2$ then $P(x,y)$ is not equal to $P(l(x),l(y))=P(x+2,y+2)$ then the relation of $y=x+1$ is not the same as $y+2=x+3$ but that just seemed wrong... any hint on the kind of automorphism I should build would be appreciated!

Answer 1

Welcome to mse!

I'm not sure why this question is so highly downvoted. It seems clear to me what you're asking, and that you have a good idea of how to attack it and have tried a few things. Regardless, here's some hints:

1. (minor hint)

Notice that $f$ preserves the decomposition $\mathbb{Z} = \text{Evens} \sqcup \text{Odds}$ in the sense that $f \! \upharpoonright_\text{Evens} : \text{Evens} \to \text{Evens}$ and $f \! \upharpoonright_\text{Odds} : \text{Odds} \to \text{Odds}$ are both bijections. So this means that the even part and the odd part don't interact! Does this encourage you to look at any particular kind of automorphism?

2. (major hint)

Since the Even and Odd parts of the structure don't interact, there are automorhpisms of the structure that do something different to even numbers and odd numbers. For instance $$\varphi(x) = \begin{cases} x & x \text{ is Even} \\ x+2 & x \text{ is Odd} \end{cases}$$ Do you see why this is an automorhpism? Do you see why it doesn't preserve $``x = y+1''$?

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I hope this helps ^_^