Could someone help me understand what is McGee's argument on restriction of consistent instances of T-schema about please?

This is gonna be messy so please bear with me.

Halbach and Holsten's "Norms for Theories of Reflexive Truth" argues that one of the desiderata of a (formal) theory of truth should be that Tr('ϕ') and ϕ (I am using 'ϕ' to mean the Godel number of ϕ here, since Mathjax doesn't work here) can be substituted salva demonstrabilitate. (I take it this means that given we can prove that ϕ is true, the two are interchangeable)

Halbach and Holsten (H&H) then talk about how we can restrict the class of instances of the T-schema.

One option is to restrict it to consistent instances, but H&H cite McGee's paper "Maximal consistent sets of instances of Tarski's Schema T" and argue that this does not suffice, and I do not understand why McGee's argument shows that.

H&H argue that McGee shows that with the diagonal lemma every sentence is equivalent to a T-schema. Thus any sentence independent from the base theory can be decided using a consistent instance of the T-schema. (Base theory refers to Peano arithmetic, or PA)

As far as I know, the diagonal lemma states that if theory T satisfies certain properties, and φ(x) is a formula with x being a free variable, then there is a sentence γ such that T⊢γ⟺φ('γ')$. But I don't see how applying this lemma gives us B_ϕ⟺(ϕ⟺ Tr('B_ϕ')).

I also don't understand what "any sentence independent from the base theory can be decided using a consistent instance of the T-schema" means, and why it shows that restricting to consistent instances does not work.

I apologise in advance of the messiness of this question, but I am really confused and I've tried my best to state it as best as I could.

H&H's argument. (Base theory refers to Peano arithmetic, or PA) enter image description here

Some preliminaries from McGee's paper, the relevant part is on what S and R mean. enter image description here

The relevant result from McGee's paper, although I suspect only the first half is relevant. (Theory S refers to some consistent arithmetical theory which entails the axioms of Robinson's R)

enter image description here enter image description here

  • Never heard of Robinson's R, are you sure it's not the usual Q? Commented Jan 3, 2022 at 5:15
  • @DoubleKnot Yea Q would definitely make more sense, but 'Robinson's R' is what McGee wrote, I can prove the screenshot of the relevant section if you would like to have a look Commented Jan 3, 2022 at 5:19
  • Sure, more is of course better... Also I remember you mentioned before in this book McGee used Yablo paradox (assuming it's a real paradox as the liar) to show such desiderata cannot allow for standard interpretation of N, so is this paragraph after that section? Anyway related? Commented Jan 3, 2022 at 5:21
  • @DoubleKnot Sure, I've added the screenshot as an edit. As for Yablo's paradox, it was Leitgeb who cited McGee, but that McGee paper is a different one and I haven't read that paper yet Commented Jan 3, 2022 at 5:27
  • 1
    Simplified the diagonal lemma says (applied to this language) that for every formula Phi(x) there is a sentence A of L_T (I assume this is the language of arithmetic plus the truth predicate) such that (A⟺ Phi(⌜A⌝)) is provable in PA. To get the B_ϕ⟺(ϕ⟺ Tr('B_ϕ')) from this you simply let Phi(x) be the formula (ϕ⟺ Tr(x)), and the result immediately follows by the diagonal lemma.
    – Johannes
    Commented Jan 3, 2022 at 13:30

1 Answer 1


The first question was answered in the comments. The short answer to the second is that they assume that the reader can figure out how the truth about there being many incompatible maximal consistent ways to extend PA with T-bivalences follows once it has been pointed out that PA proves that for every φ in L_T, there is an equivalent T-bivalence_φ, plus they give the reference that explains it in detail.

What's going on in this passage more generally:

The authors are discussing the problem faced by those wanting to develop an untyped truth theory, namely, that the theory containing PA + all instances of the T-bivalence without restriction, is inconsistent.

One proposed way to answer this (by Paul Horwich) is to include as many T-bivalence instances as possible retaining consistency (i.e. "maximal consistent" set). However the authors refer to McGee's paper that shows, for one, that there are great many such sets that are maximal consistent (with PA) but mutually incompatible. So the maximal consistency answer as such doesn't give us enough guidance.

(They also note that Cieśliński showed that restriction to a maximal conservative set suffers from similar problems.)

They point to the key fact contained in the proof that illustrates some of the problems in deciding which consistent T-bivalences to allow. As said, applying the diagonal lemma they show that for any sentence φ of L_T, PA proves that φ and some T-bivalence instance are equivalent (call this instance T-bivalence_φ).

I assume that they are trying to illustrate that while for some cases of φ, the question whether or not to include the respective T-bivalence_φ is unproblematic (e.g. when φ is inconsistent with PA, so is the T-bivalence_φ, and it should therefore not be included), in other cases the consistency criterion doesn't give us the answer.

In particular they point out that in cases where φ is independent from PA (i.e. PA doesn't prove either φ or not-φ, i.e. doesn't "decide" φ), we get a consistent extension of PA with PA+φ and with PA+not-φ, and so we also get a consistent extension by PA+T-bivalence_φ and with PA+T-bivalence_not-φ. Which of these should we choose? The section 4.1 of the SEP article that discusses this issue mentions a drastic case, a false arithmetical statement φ that is independent of PA, so there are sets of T-bivalences consistent with PA that prove false arithmetical statements.

  • I must be retarded...I can see that if there are many mutually exclusive, yet consistent sets of T-schema, this shows that restricting T-schema to consistent ones does not help us. But I just cannot for the life of me figure out how McGee/Holbach&Holsten's proof show this. I read the paragraph in H&H's paper immediately after the proof, but that only seems to be a discussion of what follows from the proof, not how one goes from the proof to the conclusion of there being mutually exclusive, yet consistent T-schema... Commented Jan 5, 2022 at 15:10
  • 1
    If sentences φ is independent from PA, then {φ} and {not-φ} are two PA-consistent sets of sentences that are mutually inconsistent. Now in the proof you can let the Δ be either one of these, and as the proof demonstrates for each one you get a maximal consistent set of T-sentences (call them Γ1 and Γ2) such that Γ1 PA-entails φ, and Γ2 PA-entails not-φ, hence, Γ1 and Γ2 are two maximal PA-consistent sets of T-sentences that are mutually inconsistent. This is enough to bring out the intuition that you can have many such incompatible maximal consistent sets of T-sentences.
    – Johannes
    Commented Jan 5, 2022 at 22:06
  • 1
    @Constantlyconfused strictly speaking you need to replace your above {B_ϕ} with {B_ϕ⟺ Tr('B_ϕ')} and {B_¬ϕ} with {B_¬ϕ⟺ Tr('B_¬ϕ')}, respectively since these are true T-sentences of Γ. It's easy to envision Γ has no PA-inconsistent sentences such as the Liar sentence, but even Γ is consistent with PA, you're right it may not be unique as reasoned above for the cases where ϕ and ¬ϕ are not provable in PA such as Con(PA). But in this case the axiomatic truth theory Γ obviously decides ϕ via its T-sentence, say decides it's false here, then it's false sentence in PA. So this is a non-conservative Commented Jan 8, 2022 at 4:41
  • 1
    @Constantly confused Pretty much. And with repeated application of the incompleteness theorem it can be shown that there are infinitely many such mutually incompatible maximal PA-consistent sets of T-sentences. As Double Knot noted SEP says there are uncountably many different sets like that. I guess this is my last comment on the matter.
    – Johannes
    Commented Jan 8, 2022 at 4:44
  • 1
    @Constantlyconfused (ctn...) extension of PA which deflationism rejects obviously which is becoming popular in the market. So any complete augmented axiomatic theory of truth with base theory incomplete is insufficient and problematic for defaltionists... Commented Jan 8, 2022 at 5:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .