Let $\mathcal{N}^{\ast}=(N, <_{\mathcal{N}^{\ast}},...)$ be a model of $Q$ (the Robinson arithmetic).
Show, that it exists a "beginning" (definition below) of $\mathcal{N}^{\ast}$ which is isomorphic to $\mathcal{N}=(\mathbb{N}, <, +,\cdot, E, 0, 1)$ ($E$ notes the exponatiation)
"beginning": A "beginning" is a model of the form $\mathcal{N}^{\ast}_{|X}$, where $X=N$ or it exists a $x\in N$ with $X=\{y\in N: y<_{\mathcal{N}^{\ast}} x\}$
I gave our definition of $Q$ here: Peano Arithmetic, proof (Axioms 1-9)
Hello,
I have a question to this task. I have to show, that $\mathcal{N}^{\ast}\cong\mathcal{N}$ hence the structures are isomorphic.
Therefore I have to give an isomorphism $\pi:N\to\mathbb{N}$ with certain properties.
But I am unsure, if there is not an easier way to solve this. $\mathcal{N}$ contains many relationsyombols, functions and constants. So it could get a bit hairy. Also I dont know if $\mathcal{N}^{\ast}$ contains the same functions, relationsymbols and contains, since only $<_{\mathcal{N}^{\ast}}$ is explicit mentioned. But I think so.
Anyways. I have to show that it exists a beginning, such that it is isomorphic to $\mathcal{N}$. So $X$ has to be infinite, otherwise I can not give a bijection.
Do you have a tip on how to solve this? Thanks in advance.