**MW:** So we have our setup: *B*⊆*M*⊆*N*, with *N* a model of PA, *B* a set of “diagonal indiscernibles” (whatever those are) in *N*, and *M* the downward closure of *B* in *N*. So *B* is cofinal in *M*, and *M* is an initial segment of *N*. I think we’re not going to go over the proof line by line; instead, we’ll zero in on interesting aspects. Where do you want to start?

**BS:** Let’s start with how we’ll end up proving *M* is a model of PA. As we do that, maybe you’ll need to bring in more about *B*, and/or about using the PHP in *N—*but since my “real” interest is more general, it would be fine if it turns out the technique only depends on a few aspects of those!

**MW:** Sounds good! Well, there are three parts to showing that *M* is a model of PA. (1) *M* is closed under + and ·. (2) *M* satisfies all the axioms other than the induction axioms. (3) *M* satisfies the induction axioms.

Parts (1) and (3) use the properties of diagonal indiscernibles. Part (2) turns out to be the easiest: it uses just the fact that *M* is an initial segment closed under + and ·. Incidentally, an initial segment closed under successor is called a *cut*, but there doesn’t seem to be a standard term for one closed under + and ·. A *supercut*?

Part (1) uses the “growth properties” of *B*. Part (3) uses that transfer principle I talked about last time, without explaining what it was. And Part (2) uses a simpler transfer principle.

It might make sense to start with Part (2), just assuming Part (1). It’s a nice warm-up.

**BS:** Ok*—*whatever order you think is best. But it seems to me that given (1) and that *M* is an initial segment containing 1, (2) is trivial. Am I missing something?

**MW:** Nope, it is pretty easy. A warm-up, like I said. I just posted the PA axioms (two equivalent versions, in fact). If you peruse them, you’ll see that apart from the induction axioms, all but one are ∀_{1} formulas: a block of universal quantifiers followed by a quantifier-free predicate. ∀_{1} formulas are *persistent downwards*. So (aside from that exceptional axiom) that takes care of Part (2) in one fell swoop!

Nothing specific to PA here—this is a general fact of model theory. If *A* is a substructure of *B*, and is an ∀_{1} formula, and for a list of elements , then . “As above, so below.” (Note that we must demand , otherwise wouldn’t make sense.)

Our first transfer principle, nice and easy. We need Part (1), by the way, because without it we don’t know that *M* is a substructure of *N*.

**BS:** In hindsight, it’s well worth seeing that spelled out—it’s straightforward, but not as trivial as I thought. I do see how to take care of the “holdout”:

∀*x*∀*y* ( *x*<*y* ↔ ∃*w x*+(*w*+1)=*y* )

We have to prove that if *x* + *w* + 1 = *y*, then *w* < *y*, to make sure *w* is in *M*. As easy as that is, I hadn’t explicitly noticed the need to prove it, until you pointed out how that axiom differs from the others.

**MW:** Right. The trick is to break it into two axioms:

∀*x*∀*y* ( *x*<*y* → ∃*w x*+(*w*+1)=*y* )

∀*x*∀*y* ( ∃*w x*+(*w*+1)=*y *→ *x*<*y* )

Because *P*→*Q* is the same as ¬*P*∨*Q*, the second half is *also* ∀_{1}. For the first half, we have the key fact that you noted: for *x,y*∈*M*, the *w* given by the axiom is less than *y*, so it also belongs to *M*.

So another transfer principle handles the axiom: Π_{1} downwards persistence. Namely: say we have two structures for L(PA), *A* and *B*, with *A*⊆*B*. Suppose also that *A* is an initial segment of *B*. Then Δ_{0} formulas are absolute between *A* and *B*: persistent both upwards and downwards. And Π_{1} formulas are persistent downwards. The initial segment requirement has paid off by making more formulas persistent.

Now, “∀*x*∀*y*(*x*<*y* → ∃*w x*+(*w*+1)=*y*)” is ∀_{2}, aka ∀∃. We can’t apply any of these transfer principles to it. But it’s implied by this formula: “∀*x*∀*y*(*x*<*y* → ∃*w<y x*+(*w*+1)=*y*)”. That’s a Π_{1} formula. It holds in *N*, so it holds in *M*.

**BS:** That’s interesting! I guess it is worth reformulating as many axioms as possible into Π_{1} form, to make it clearer that they transfer this way.

**MW:** More generally, this shows the insights afforded by knowing where formulas land in the arithmetic hierarchy.

OK, let’s backtrack to Part (1)*—*showing that *M* is closed under + and ·. A couple of observations first. Suppose *r*_{1}*<r*_{2}*<…<r _{n}*<… is an infinite increasing sequence of elements of

*N*, and

*M*is the downward closure of this sequence. If

*r*

_{n}_{+1}≥2

*r*for all

_{n}*n*, then

*M*is closed under addition. If

*r*

_{n}_{+1}≥

*r*

_{n}^{2}for all

*n*, then

*M*is closed under multiplication.

**BS:** Because everything in *M* has to be under *some* *r _{n}*! Clever!

**MW:** Now we need the definition of a *set of diagonal indiscernibles* in the model *N* for a set Φ of formulas. Here’s what I said in post 9:

For diagonal indiscernibles, the truth-value of φ(

r_{1},…,r,_{n}c_{1},…,c) is the same no matter what the_{h}r’s,_{i}providedthat {r_{1},…,r} is a subset of_{n}B, thatr_{1}<…<r, and that all the_{n}c’s are less than an indiscernible_{i}bwhich is less than all ther’s. (Think of_{i}bas a kind ofbarrier—or DMZ.) And of course that φ belongs to the given set Φ of formulas—Δ_{0 }formulas in this case.

**BS: **I’m not sure I understand that correctly—is the following reformulation equivalent?

Given φ and *c*_{1},…,*c _{h}*, let

*b*be minimal in

*B*such that

*c*

_{1}

*,…,c*<

_{h}*b*. Then for every

*r*

_{1}

*<…<r*all greater than

_{n}*b*, φ(

*c*

_{1},…,

*c*,

_{h}*r*

_{1},…,

*r*) has the same truth value. (I.e., given those conditions, its truth value depends only on φ and

_{n}*c*

_{1},…,

*c*. BTW, I listed the

_{h}*c*’s first within φ, since I find it easier to understand that way.)

_{i}**MW:** I prefer not to assume that the barrier is minimal. It’s enough that it separates the parameters from the other indiscernibles. Let me try another phrasing. I’ll also change notation just a bit, for later convenience.

*B*⊆*N* is a set of diagonal indiscernibles for Φ in *N*, if the following holds. For every formula φ(*x*_{1}*,…,x*_{h}, *y*_{1},…,*y _{n}*) in Φ and for every tuple of elements

*c*

_{1}

*,…,c*in

_{h}*N*and for every

*b*,

*b*

_{1}

*,…,b*,

_{n}*e*

_{1}

*,…,e*∈

_{n}*B*, if

*c*_{1}*,…,c _{h}*<

*b*<

*b*

_{1}

*<…<b*and

_{n}*c*

_{1}

*,…,c*<

_{h}*b*<

*e*

_{1}

*<…<e*

_{n}then

*N*⊧φ(*c*_{1}*,…,c _{h}*,

*b*

_{1}

*,…,b*

_{n}) ⇔

*N*⊧φ(

*c*

_{1}

*,…,c*,

_{h}*e*

_{1}

*,…,e*

_{n}).

I call *b* the *barrier*, and everyone calls the *c*_{i}’s the *parameters*. I’m using the *b*_{i}’s and *e*_{i}’s for arbitrary increasing sequences of indiscernibles, unlike the *r*_{i}’s: they’re supposed to be *all* the indiscernibles, so *B*={*r*_{1}, *r*_{2}, *r*_{3},…}.

The numbers *h* and *n* will depend on the formula φ. Notice that the *b _{i}*’s and the

*e*’s must be in increasing order, but the

_{i}*c*’s don’t have to be. As for the order of

_{i}*b*’s and

_{i}*e*’s with respect to each other, nothing is assumed about that.

_{i}**BS: **Ok, that’s clear. How does the barrier end up helping us?

**MW:** The key fact is that the barrier must belong to *B*; it turns out that any increasing sequence of diagonal indiscernibles grows at a *very* rapid clip. So having an element of *B* in between the *c _{i}*’s on one side, and the

*b*’s and the

_{i}*e*’s on the other, means the two sides must be separated by a pretty big gap! If I may speak metaphorically, the

_{i}*b*’s and the

_{i}*e*’s are so far over the horizon of the

_{i}*c*’s, that the

_{i}*c*’s can’t distinguish between them—they’re indiscernible to the

_{i}*c*’s!

_{i}The role of the barrier may become clearer when we go through the proof that *r _{n}*

_{+1}≥2

*r*and

_{n}*r*

_{n}_{+1}≥

*r*

_{n}^{2}for all

*n*.

**BS: **Ok, I’m looking forward to that! (I might even think I can guess something about it, but I’ll wait and see.)