MW: Recap: we showed that PAT implies ΦT, where ΦT is the set of all formulas
Now we have to show the converse, that PA+ΦT implies PAT. But first let’s wave our hands, hopefully shaking off some intuition, like a dog shaking off water.
Below is a diagrammatic description of PAT. The outer rectangle contains all models of PA, the inner oval contains all those that are (isomorphic to) ω’s of models of T. In other words, all models of PAT. Now let’s say φ is a sentence that cannot be proved using only PA, but can be proved from T. So for T=ZF, φ could be the Paris-Harrington principle, or Goodstein’s theorem, or Con(PA). The blue region represents all the models where φ holds. So it includes the oval, but not all of the rectangle. The oval is the intersection of all such blue regions, as φ ranges over all of PAT. If φ were provable from PA, then it would paint the entire rectangle blue.
Next I want to take a second look at ΦT. I’m going to turn its formulas inside-out and upside-down. Start with
The contrapositive is
Basic results from proposition logic tells us that for any sentence φ and any theory T, ¬Con(T+φ) is equivalent to T⊢¬φ. (“Proof by contradiction”.) So we have
Since φ ranges over all sentences of the language of PA, we might as well replace ¬φ with φ. Upshot: ΦT is equivalent to
Let me explain the quote marks. Provability is a syntactic property, an assertion about strings of symbols. Or maybe I should say strings of strings of symbols. We can code syntactic claims into PA; the quoted stuff stands for the formalization of that claim into the language of PA. Kind of like the corner brackets for Gödel numbers, but more general.
The sentences of ΦT say “Trust T.” In other words, “If T proves it, then it’s true.”
Let’s take a second look at the implication . We want to show that the formula “paints the oval blue”, i.e., holds in all models of T. Well, if , then “it stands to reason” that φ should hold in the of a model of T. That’s the intuition, at least.
As before, some subtleties lurk, but not enough to derail the conclusion. We must interpret the assumption “” inside the model . The proof of could have non-standard length! The trick is to bring the whole intuitive argument inside the model of T, which we can do thanks to the items mentioned last time: the Reflection Principle of ZF, and the existence of a formal truth predicate for formulas with up to d quantifiers (for any given d).
The reverse implication, , takes less work, though there is one fine point. Suppose T yields . Then some finite fragment does, say
In fact (this is the fine point),
Why? Well, if Tn proves something, then PA can prove that it proves it. People often argue (correctly) that if the Goldbach conjecture is false, then it’s provably false in PA: if we have a counterexample, we can check that it is a counterexample with a finite calculation, which we can code into PA. The same holds for . Technically this is known as the Σ1-completeness of PA. Both “Tn proves φ” and “the Goldbach conjecture is false” are Σ1 statements, unlike “there are infinitely many prime pairs”, which is Π2. It’s quite conceivable that the prime pair conjecture could go either way without PA having a proof.
As we saw earlier, ΦT is equivalent to the collection of all statements of this form:
So in PA+ΦT, we can prove both and , for any φ in PAT. Modus ponens takes it from there!