In his book "Models of Peano Arithmetic" Kaye proves the Paris-Harrington Theorem. He starts off by introducing a "simplification" then proves the short Lemma 14.11 about it (see picture). I don't understand any of this. Specifically:
- What is the rational of the "Truth definition for $I$" (just above Lemma 14.11 in the picture)?
- Why does "$I \subset_e M$ is a structure of the language $\mathcal{L}_A$ of arithmetic" imply that $I \models \text{PA}^{-}$?
- I don't understand at all why the proof shows that $I \models \text{PA}$. Why does the proof imply that $I$ models all induction axioms?
Basically, can someone explain the proof and its background, because I'm completely lost. I would greatly appreciate any help!
Re: (2), first remember that all but one of the axioms of $\mathsf{PA^-}$ are universal, hence downwards-absolute between structures: since $I$ is a substructure of a model of $\mathsf{PA^-}$ it satisfies all of these sentences too.
So this just leaves the thirteenth axiom: $$\mathsf{Ax13}:\quad\forall x,y(x<y\rightarrow \exists z(x+z=y)).$$
This is where we use the fact that $I\subset_eM$, or more intuitively that $I$ is downwards closed. Let $x,y\in I$ with $x<y$. Applying $\mathsf{Ax13}$ in $M$ we get in $M$ some $z$ such that $x+z=y$. In $M$ we have $z\le y$; since $I$ is downwards closed this means $z\in M$. Now the property $x+z=y$ is absolute between $I$ and $M$, so $I\models x+z=y$. And this gives $\mathsf{Ax13}$ in $I$ as desired.
Note that we needed end extension as opposed to mere superstructurehood, in verifying axiom $13$. It's a good exercise to check that every nonstandard model of $\mathsf{PA}$ has a (necessarily non-downwards-closed) substructure not satisfying axiom $13$. On the other hand, if we included "symmetric subtraction" $(x,y)\mapsto \vert x-y\vert$ as a primitive operation, then $\mathsf{Ax13}$ can be rephrased as a universal and after such rephrasing every substructure of a model of $\mathsf{PA^-}$ would again satisfy $\mathsf{PA}^-$.