In my proof theory monograph (proof theory and logical complexity, Girard from '87) there is an exercise 1.5.4. on page 78 called 'Tarski's theorem'.

It says:

"Show that there is no truth predicate for $L_0$ in PA, i.e. no formula TR such that,
if A is any closed formula of $L_0$,
then $PA \vdash TR[\overline{\ulcorner A \urcorner}] \leftrightarrow A $."

($L_0$ is the language of elementary arithmetic.)

I was suprised: Is this really a formulation of Tarski's theorem, or does it follow immediately from it (or maybe from Gödel's theorems, in this formulation)?

Isn't Tarski's theorem dealing with $truth$ instead of provability?
There is no $L_0$-definable predicate (or formula, in this case) TR such that, for every closed formula A:

TR$(\ulcorner A \urcorner)$ is true in PA
if and only if
$A$ is true in PA.

Wouldn't the above formulation from the exercise ask for a formula defining provability rather than truth?

cheers, Ettore


Tarski's theorem is about the undefinabilty of a truth predicate in a model.

To take a concrete example, let $\mathfrak{N}$ be the standard model for $\mathbf{PA}$. Then a relation $R(x_1,...,x_k)$ is definable in $\mathfrak{N}$ if there is a (first order) formula $\varphi(x_1,...,x_k)$ in $L_0$ such that, for all $n_1,..., n_k \in \mathbb{N}$, $$ R(n_1,...,n_k) \iff \mathfrak{N} \vDash \varphi(\bar{n}_1,...,\bar{n}_k).$$

Then Tarski's Theorem says that there is no formula $TR$ in $L_0$ such that $$\text{$\varphi$ is true} \iff \mathfrak{N} \vDash TR(\ulcorner\varphi\urcorner).$$ This is a semantic notion: for models. But it follows from the impossibility of a formula $TR$ such that $\mathbf{PA} \vdash \varphi \leftrightarrow TR(\ulcorner\varphi\urcorner)$ by the Diagonal Lemma:

Diagonal Lemma (for PA) If $\psi(x)$ is a formula of $L_0$ with $x$ free, then there is a sentence $\varphi$ in $L_0$ such that $$\mathbf{PA} \vdash \varphi \leftrightarrow \psi(\ulcorner\varphi\urcorner).$$

We can use $\lnot TR(x)$ instead of $\psi(x)$ above to get a sentence $\lambda_{\mathbf{PA}}$ (for liar!) such that $$\mathbf{PA} \vdash \lambda_{\mathbf{PA}} \leftrightarrow \lnot TR(\ulcorner\lambda_{\mathbf{PA}}\urcorner).$$

But $TR$ is supposed to satisfy $$\mathbf{PA} \vdash \lambda_{\mathbf{PA}} \leftrightarrow TR(\ulcorner\lambda_{\mathbf{PA}}\urcorner),$$ which clearly yields a contradiction with the statement above.

Hence, assuming $\mathbf{PA}$ is consistent, there is no formula $TR$ in $L_0$ such that $\mathbf{PA} \vdash \varphi \leftrightarrow TR(\ulcorner\varphi\urcorner)$.

Furthermore, if $\mathbf{PA}$ is sound, it follows from $\mathbf{PA} \vdash \lambda_{\mathbf{PA}} \leftrightarrow \lnot TR(\ulcorner\lambda_{\mathbf{PA}}\urcorner)$ that $$ \mathfrak{N} \vDash \lambda_{\mathbf{PA}}\leftrightarrow \lnot TR(\ulcorner\lambda_{\mathbf{PA}}\urcorner),$$

and therefore

$$\mathfrak{N}\vDash\lambda_{\mathbf{PA}} \iff \mathfrak{N}\vDash\lnot TR(\ulcorner\lambda_{\mathbf{PA}}\urcorner),$$

so clearly $TR$ does not define "truth" in $\mathfrak{N}$!

The Diagonal Lemma relies on the fact that $\mathbf{PA}$ is "strong enough" to represent all computable relations (this works already for $\mathbf{Q}$, and $\mathbf{Q}\subset\mathbf{PA}$).

Addendum Truth and Proof

I'd like to further clarify your question of why this $TR$ predicate in $\mathbf{PA}$ is about truth and not provability.

In formal languages, sentences are considered just collections of symbols, they don't mean anything in themselves until given an interpretation (semantics). We can also stipulate rules to infer sentences from other sentences purely formally or syntactically (proof theory).

Now we have a property of sentences, "truth", which we want to capture in our formal language, $L_0$ through a sentence, $TR$. We then expect this sentence to "behave" properly. But we can ask this question in two ways:

  • Semantics. Does the interpretation (the models) respond correctly to this predicate? In this case we want definability: $$\text{$\varphi$ is true} \iff \mathfrak{N} \vDash TR(\ulcorner\varphi\urcorner).$$

  • Proof Theory. Does the formal system (with axioms and rules) respond correctly to this predicate? This is representability and is slightly more involved:

    1. If $\varphi$ is true, then $\mathbf{PA} \vdash TR(\ulcorner\varphi\urcorner).$
    2. If $\varphi$ is false, then $\mathbf{PA} \vdash \lnot TR(\ulcorner\varphi\urcorner).$

In both cases, the predicate $TR$ is supposed to capture when a sentence is "true", but it does so in different contexts (can the model tell? can we prove it formally?).

Since $\mathbf{PA}$ is a formal system, it doesn't make sense to say "$\varphi$ is true in $\mathbf{PA}$" unless you have in mind a model; $\mathbf{PA}$ is just a collection of axioms (or axioms plus its derivable consequences, depending on the definition).

We could ask the same thing about a provability predicate $PRV$: does the model recognise the predicate appropriately? Does $\mathbf{PA}$ recognise the predicate appropriately? Gödel's Incompleteness Theorems rely on the fact that $PRV$ can be correctly represented in $\mathbf{PA}$ (assuming consistency):

  1. If $\mathbf{PA} \vdash \varphi$, then $\mathbf{PA} \vdash PRV(\ulcorner\varphi\urcorner)$.
  2. If $\mathbf{PA} \not\vdash \varphi$, then $\mathbf{PA} \vdash \lnot PRV(\ulcorner\varphi\urcorner)$.

Addendum II Negations inside and outside the system

There are two notions of "negation":

  • $\mathfrak{N}\vDash\lnot\varphi$: the sentence $\lnot\varphi$ is true is the model;
  • $\mathfrak{N}\not\vDash\varphi$: the sentence $\varphi$ is false in the model.

Naturally, you want the two to be equivalent, so we stipulate:

$$\mathfrak{N}\vDash\lnot\varphi \iff \mathfrak{N}\not\vDash\varphi$$.

This is fine because models are supposed to be complete: they give an interpretation to every sentence of the language, so either $\mathfrak{N}\vDash\varphi$ or $\mathfrak{N}\not\vDash\varphi$; hence $\mathfrak{N}\vDash\varphi$ or $\mathfrak{N}\vDash\lnot\varphi$.

But this is different in a formal system, say $\mathbf{PA}$. Compare:

  • $\mathbf{PA}\not\vdash\varphi$: $\mathbf{PA}$ doesn't prove $\varphi$;
  • $\mathbf{PA}\vdash\lnot\varphi$: $\mathbf{PA}$ proves $\lnot\varphi$.

As I suppose you know from Gödel's First Incompleteness Theorem, $\mathbf{PA}$ is incomplete: there is a sentence $\gamma_\mathbf{PA}$ such that $\mathbf{PA}\not\vdash\gamma_\mathbf{PA}$ and $\mathbf{PA}\not\vdash\lnot\gamma_\mathbf{PA}$, so it is certainly not the case that $$\mathbf{PA}\vdash\lnot\varphi \iff \mathbf{PA}\not\vdash\varphi.$$

This is why, when we define representability (say of $PRV$) in $\mathbf{PA}$ we need two clauses (1) and (2) at the end of the previous addendum. Merely stipulating

$$\mathbf{PA}\vdash\varphi \iff \mathbf{PA} \vdash PRV(\ulcorner\varphi\urcorner)$$

would be too weak: $\mathbf{PA}\not\vdash\varphi$ would only imply $\mathbf{PA}\not\vdash PRV(\ulcorner\varphi\urcorner)$ ("if we can't prove $\varphi$, then we can't prove that it can be proved"), which is not the same as $\mathbf{PA} \vdash \lnot PRV(\ulcorner\varphi\urcorner)$ ("if we can't prove $\varphi$, then we can prove that it can't be proved").

  • $\begingroup$ Hey Ryan thanks for your great answer! That's clarifying. However...not being really familiar with model theory (I only studied a common logic textbook a few years ago and am now studying a bit proof theory on my own), I wonder whether the part between 'Since models are...' and 'It follows that..' is really necessary, cause: Doesn't $$ \mathfrak{N} \vDash \varphi\leftrightarrow TR(\ulcorner\varphi\urcorner).$$ already mean your last conclusion, that $\varphi$ is true if and only if $TR(\varphi)$ is true? Not that what comes inbetween seems wrong to me or something, but somehow 'detoury'... $\endgroup$ – Ettore Dec 5 '18 at 14:34
  • 1
    $\begingroup$ Hi @Ettore. I've added a proof-theoretic version of the proof using the Diagonal Lemma (although this assumes consistency: if $\mathbf{PA}$ is inconsistent, then it proves anything you like!). Maybe that's what you were after? $\endgroup$ – ryan221b Dec 5 '18 at 16:30
  • 1
    $\begingroup$ As four the "detour": whether $\mathfrak{M}\vDash\varphi$ or not depends on what $\mathfrak{M}$ takes $\varphi$ to mean. In the case of arithmetic, $\varphi$ is interpreted as a sentence about natural numbers and $\mathfrak{N}$ is the "intended" interpretation. So, "by design", $\varphi$ is true if and only if $\mathfrak{N}\vDash\varphi$. But if you chose a different model, that may no longer be the case. $\endgroup$ – ryan221b Dec 5 '18 at 16:38
  • 1
    $\begingroup$ For example, as an arithmetical statement, $\forall x \forall y (x + y = y + x)$ is "true", but there are models $\mathfrak{M}$ of $\mathbf{Q}$ such that $\mathfrak{M} \not\vDash \forall x \forall y (x + y = y + x)$. $\endgroup$ – ryan221b Dec 5 '18 at 16:38
  • 1
    $\begingroup$ You're welcome :) I hope my second addendum clarifies the difference between negating $\vDash$ and $\vdash$, and "$\lnot$". I have also rephrased Tarski's Theorem: you use the "proof theoretic" version to deduce the model theoretic conclusion. As for derivability conditions, if you mean the Hilbert-Bernays ones, I think that would be better suited for a new question. $\endgroup$ – ryan221b Dec 6 '18 at 11:53

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.