Proving Undecidability of first order logic without first proving it for arithmetic.

Question

All text I have read prove the Undecidability of first order logic a bit as an afterthought and after having proved the incompleteness and Undecidability of (Peano) Arithmetic.

This proof also hardly works in philosophy, I would like a proof that first order logic is undecidable, but without using arithmetic , Godels incompleteness theorems and all of that.

It is just such a tortuous route and get many will lost halfway, it is about first order logic why do we need to add arithmetic to proof undecidability?

My own first ideas of a proof:

(not sure even if it is correct, I posted this earlier in little different form at , Gödel's Completeness Theorem and satisfiability of a formula in first-order logic , in the hope of comments)

• There are some formula's where the answer about validity depends on the size of the domain.

$\exists x R(x) \to \forall x R(x) $ is true when the domain only contains one element, but false in any larger domain.

Suppose now we have a formula $\phi $ that is false in every finite domain but is true in an infinite domain.

• We cannot construct a model of $\phi $ just because it requires an infinite domain.
• Also we cannot proof $\lnot \phi $ because it is just not a true statement

So by soundness $ \not \vdash \lnot \phi $ and also $ \phi $ has no finite (constructable) model so $ \phi $ and $\lnot \phi $ are undecidable statements

And for $\phi $ we can just use $\lnot ( \forall x \forall y \forall z (((Rxy \land Ryz) \to Rxz ) \land \lnot Rxx) \land \forall y \exists x Rxy )$

I agree $\phi $ is a kind of (minimal) arithmetic in disguise, interprete Rxy as $ x > y $ , but it is not full arithmetic and the proof is much shorter.

But is this a correct line of reasoning?

Has a model to be constructive? ($ \phi $ is "obviously" satisfiable it is just not constructable)

Are there other and better ways to explain it? (references are very welcome)

Answer 1

I would like a proof that first order logic is undecidable, but without using arithmetic

But hold on! The proposition "first order logic is undecidable" is, when unpacked, just a proposition of the form "there is no computable function which does so-and-so". So of course a proof of this proposition is going to have to come from an application of the general theory of computable functions. (We might only need a naive ability to recognise a computation when we see it to establish the positive claim that some function can be computed: but to establish a negative, that there is no computable function which does so-and-so, we do need some general theory which can enable us to talk about the limits of the computable).

Now, the relevant computations here are over finite strings that constitute wffs and proofs, so the background theory we need in fact is or is tantamount to the theory of arithmetical computable functions (for you can code finite strings by numbers in familiar ways; or indeed we can go in the opposite direction and reconstruct arithmetic in the theory of finite strings). So the natural lines of proof that first order logic is undecidable will come from the theory of computable arithmetical functions (perhaps in slight disguise). In sum, broadly arithmetical considerations (far from being extraneous, a "tortuous" diversion) are exactly what we would expect to use here.

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Added: What I wrote above was [at least intended to be] consistent with the sort of points Carl Mummert's makes in his typically elegant answer. It is the theory of computability that needs to be invoked: but the point at which it gets applied doesn't have to be to a formal arithmetic. As Carl very nicely points out, we could go instead via Tarksi's proof about groups. However, I was taking it that that sort of thing would seem to the OP even more "tortuous" a route to the undecidability result; and we are still applying the theory of computable numerical functions.

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That's the main point to be made, but I can perhaps add a subsidiary comment on the remark

We cannot construct a model of ϕ just because it requires an infinite domain.

Well, by any ordinary constructive standards the wff you describe (losing the unintended negation) has a constructively acceptable model -- the natural numbers in their natural order! What is wrong with that?

Answer 2

Yes, it is possible to prove the undecidability of first-order logic without formal arithmetic. Undecidability is mostly about computability theory, in the end.

For example, Tarski proved that the theory of groups is undecidable: there is no procedure to decide if a formula $\phi$ in the language of groups is provable from the group axioms. But the group axioms are finite in number, so we can write them as a single axiom $G$. Then it follows from Tarski's result that there must be no procedure to determine whether a sentence of the form $G \to \phi$ is provable (with no extra axioms), when $\phi$ is in the language of groups. Thus, in general, first-order provability is not decidable.

Tarski's proof works by showing that the the theory of groups can be used to encode facts about computation, so that deciding which sentences about groups are provable would allow us to solve the halting problem. But Tarski's proof does not require anything about formal arithmetic.

What we get from formal arithmetic is a stronger property. Theories such as Peano arithmetic are essentially undecidable: not only is Peano arithmetic undecidable, but any consistent, effectively axiomatized extension of Peano arithmetic is also undecidable. This is not the case for the theory of groups. If we add $(\forall x)(\forall y)[x=y]$ to the group axioms, we obtain the theory of a one-element group, which is decidable.

Answer 3

The book Computability and Logic by Boolos and Jeffrey relates the undecidability of first-order logic directly to the halting problem: by providing a first-order logic description of any turing machine and input, the halting problem can be stated as a first-order logic argument. Thus, if first-order logic is decidable, the halting problem is solvable. Since it is not, first-order logic is not decidable.

Answer 4

Actually, it should be possible to proof the undecidability of first-order logic without any background theory, so just in plain logic, by reduction to the Halting Problem (actually, Turing and Church did this, in my opinion). Have a look at this web page from Stanford university.

Answer 5

I would like to add some reflections about the "non constructivity" issue, relating some previous posts about Gödel's Completeness Theorem and satisfiability and Mathematical content of Gödel's Completeness Theorem and Roy Simpson Constructiveness of Gödel's Completeness Theorem .

My reflections are the following :

(i) Modern proofs of Gödel's Theorem

I think that those in the Gentzen's tradition are more "in the spirit" of the original G's proof.

References to proof theory textbooks :

• Kurt Schutte, Proof Theory (engl transl 1977 - 1st german ed 1960), see page 29,

• Gaisi Takeuti, Proof Theory (2nd ed 1987 - 1st ed 1975), see page 40,

• Sara Negri & Jan von Plato, Structural Proof Theory (2001), see page 81,

and my "beloved" : S.C.Kleene, Mathematical Logic (1967) and R.Smullyan, First-Order Logic (1969).

First of all, they clearly relate the theorem to a proof systems (this is my "very very personal" feeling: I do not like proofs that validate the Theorem without any mention to a proof system).

Second, due to "hilbertian origin" of proof theory, they are very sensitive at declaring the "mathematical resources" needed in the proof (König's lemma); see Schutte, page 30.

The proof [...] is non-constructive since it draws on the non-constructive notions of validity and König's lemma, which has a non-constructive proof.

Third, in the construction of the proof, they build a tree that is

constructively defined and it follows from the proof that for every deducible formula one have a standard deduction which is given, constructively, by the [tree].

again from the Remark in Schutte,'s textbook, page 32.

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(ii) Validity and infinite domains

Assuming that the above reflections are correct, I will add, following Peter Smith's note about Squeezing arguments (2010), that the above proofs clearly relate the notions of "validity-in-virtue-of-form", as defined during the '20s in the Hilbert school, from Bernays to Gödel's 1930 paper, with the notion of "derivability-in-a-proof-system" and with the notion of "having a countermodel in the natural numbers", all of which are the "basic ingredients" of Gödel's original proof.

Having said that, I'm asking about the "ontological committment" of the above proof.

May we dispense with $\mathsf {ZFC}$ ?

We need really only König's lemma, natural numbers, induction.

It is enough some weak subsystems of second-order arithmetic in order to formalize this piece of mathematics ?

According to some previous answer by Carl Mummert, $\mathsf {WKL_0}$ is enough.

According to the OP's words "the answer about validity depends on the size of the domain": true. But if we leave the infinite domain of natural numbers (i.e.$\mathbb {N}$) we loose any "interest" in predicate logic.

If I assume a limitation to finite domains, I can dispense altogether with quantifiers and stay with finite conjunctions and disjunctions; so propositional logic is enough, and we have decidability and effective methods (like truth-table) for checking validity.