Curry’s paradox in Turing machine land?

Question

Background

There’s an excellent question on MathOverflow that talks about the following: imagine that $M$ is a TM that searches over all ZFC proofs and halts if and only if it finds a proof that it halts. The answer is yes because it’s possible to adapt the logic behind Löb’s theorem to Turing machines.

At that question is a remarkable answer by Akiva Weinberger that spells out an explicit argument, based on the proof of Löb’s theorem, showing why this machine $M$ halts:

Build a second machine $N$. $N$ searches for a proof in ZFC of, "if $N$ halts then $M$ halts". If it finds one, it halts.

ZFC can argue as follows. "Suppose $N$ halts. Then it found a proof that if $N$ halts then $M$ halts. This, combined with the trace of $N$ halting, would be a ZFC proof that $M$ halts. Thus $M$ finds this and halts."

That paragraph is a ZFC proof that if $N$ halts then $M$ halts… which $N$ finds, so $N$ halts. Thus, by the same logic as above, there is a proof that $M$ halts, which it will find and then halt on.

Löb’s theorem is closely related to Curry’s paradox. Briefly, Curry’s paradox envisions a statement $P$ that says “if $P$ is true, then $\phi$ is true,” where $\phi$ is some arbitrary statement. The paradox is that this seems to imply that $\phi$ is true regardless of what $\phi$ is.

My Question

Pick any statement $\phi$. I’m going to argue that ZFC proves $\phi$. (Clearly it doesn’t, so there has to be a flaw in the following reasoning. However, I’m not sure what it is!) This is essentially a minimally-modified version of Akiva Weinberger’s MO answer but applied to Curry’s paradox rather than Löb’s theorem. To make it easier to see the changes, I’ve bolded all parts that differ from the original.

Consider a TM $M$ that searches over all ZFC proofs and halts if and only if it finds a proof of the statement “if $M$ halts, then $\phi$ is true.”

Build a second machine $N$. $N$ searches for a proof in ZFC of, "if $N$ halts, then if $M$ halts, then $\phi$ is true.” If it finds one, it halts.

ZFC can argue as follows. "Suppose $N$ halts. Then it found a proof that if $N$ halts, then if $M$ halts, then $\phi$ is true. This, combined with the trace of $N$ halting, would be a ZFC proof that if $M$ halts, then $\phi$ is true. Thus $M$ finds this and halts. This, combined with a trace of $M$ halting, would be a ZFC proof of $\phi$.

That paragraph is a ZFC proof that if $N$ halts, then if $M$ halts, then $\phi$ is true… which $N$ finds, so $N$ halts. Thus, by the same logic as above, there is a proof that $M$ halts, which it will find and then halt on. And since $M$ halts, a trace of $M$ halting, plus the preceding argument, gives a ZFC proof of $\phi$.

There has to be something wrong with the above argument, but I confess I’m not seeing what it is. Can someone point out where I’m telling lies? :-)

Answer 1

I don't see a contradiction anywhere. Let $\varphi$ be a statement in the language of $\mathsf{ZFC}$. Let $\ulcorner\varphi\urcorner$ be a string encoding $\varphi$. Let $\text{prov}(\ulcorner\varphi\urcorner)$ be the formula that says there exists a string encoding a proof of $\varphi$ from the axioms of $\mathsf{ZFC}$.

For a Turing machine $M$ and a string $\ulcorner M\urcorner$ encoding $M$, let $\text{halt}(\ulcorner M\urcorner)$ be the formula that says there exists a string encoding a run of $M$ halting on the empty input.

What you showed is that there is a Turing machine $M_\varphi$, encoded by a string $\ulcorner M_\varphi\urcorner$ such that $$\mathsf{ZFC}\vdash\text{halt}(\ulcorner M_\varphi\urcorner)\rightarrow\text{prov}(\ulcorner\varphi\urcorner)$$

and also that there is a Turing machine $N_\varphi$ encoded by $\ulcorner N_\varphi\urcorner$ such that $$\mathsf{ZFC}\vdash(\text{halt}(\ulcorner N_\varphi\urcorner)\land\text{halt}(\ulcorner M_\varphi\urcorner))\rightarrow\text{prov}(\ulcorner\varphi\urcorner)$$

but unless $\mathsf{ZFC}\vdash(\text{halt}(\ulcorner N_\varphi\urcorner)\land\text{halt}(\ulcorner M_\varphi\urcorner))$ or $\mathsf{ZFC}\vdash\text{halt}(\ulcorner M_\varphi\urcorner)$ you cannot conclude from this that $\mathsf{ZFC}\vdash\text{prov}(\ulcorner\varphi\urcorner)$.

It's difficult to understand why you need either $M$ or $N$. You can build a machine $M$ that searches the space of strings for proofs of $\varphi$ in $\mathsf{ZFC}$ directly. Why do you need a machine $M$ to search for proofs that 'If $M$ halts on the empty input then $\varphi$ is provable in $\mathsf{ZFC}$'? A machine that searches for proofs of $\varphi$ will halt on the empty input if and only if $\varphi$ is provable in $\mathsf{ZFC}$.

Answer 2

The problem is this part:

That paragraph is a ZFC proof that if N halts, then if M halts, then ϕ is true

The relevant paragraph only talks about proofs of $\phi,$ not $\phi$ itself. So all you can say at that point is:

That paragraph is a ZFC proof that if N halts, then if M halts, then ZFC proves ϕ

which isn't as useful.

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A good way to find the mistake is to take $\phi=\bot$ (i.e. any contradiction). Then using the 2nd incompleteness theorem it's easy to see that these are equivalent in a weak metatheory:

• $M$ halts
• $N$ halts
• ZFC is inconsistent

In particular we (and ZFC) can prove "if $M$ and $N$ halt then $\mathsf{ZFC}\vdash \phi$". But ZFC cannot prove "if $M$ and $N$ halt then $\phi$", unless it's inconsistent.

Answer 3

I think I have an answer. The slogan is: the subtle assumption we're making is precisely the "Löb hypothesis", and so being careful about this assumption, we get directly a proof of Löb's theorem using these funny Turing machine.

Let me analyze a simpler variant, that makes the point just as well. Let $P$ denote some proposition (like $P$ is "$0=1$", or "$\forall x \forall y (x+y=y+x)$").

Consider the Turing machine/program $N$ that halts iff it finds a (PA/ZFC/whatever) proof that [$N$ halts $\to P$] (a proof which it then prints out).

Side notes: this means a proof that can use "$N$ halts" (encoded in PA/ZFC/whatever; indeed PA and ZFC are sufficiently powerful languages that they can encode this) as a black box wherever/whenever it wants in a proof tree that concludes with "$P$".

From now on, let me just say "PA" instead of "PA/ZFC/whatever".

Also one can construct this $N$ by: defining $N_1$ to be a 1-input program that takes in the code $\texttt{B}$ for a 1-input program $B$ and runs through PA proofs until it finds a PA proof that [$B$ halts $\to P$]; and then define $N$ to be $N_1(\texttt{N}_1)$.

Back to the argument at hand:

Let us assume that $N$ halts (like truly halts, in the true real world).

Because halting is something that can be "finitely witnessed", we know that PA can prove [$N$ halts].

Then by construction of $N$, it prints out a PA proof that [$N$ halts $\to$ P].

Putting these together, we can conclude that [PA proves $P$]. (this is all under the initial assumption at the start of this textblock)

The above argument can be formalized in PA (I think), so $(\star)$: we now have a PA proof that [$N$ halts $\to$ [PA proves $P$]].

Assuming the Löb hypothesis (!!!), i.e. PA proves [$\operatorname{Prov}_{\text{PA}}(P)\to P$], only then can we chain implications together to simplify $(\star)$ to just: we have a PA proof that [$N$ halts $\to P$].

This leads to $N$ halting (truly, in the true real world), which again PA can see/prove, and so applying modus ponens one last time we arrive at [PA proves $P$].

(I agree with the MO post --- this must be funniest proof I've ever seen. The necessary "double lap", i.e. doing essentially the same argument twice, is absolutely hilarious.)

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One thing I think is super super subtle about the Löb hypothesis is that it's actually saying $\{\text{PA}, \operatorname{Prov}_{\text{PA}}(P)\} \vdash P$, where can only use $\operatorname{Prov}_{\text{PA}}(P)$ as a black box in the proof tree.

Whereas if we assume PA is consistent/honest (which most people believe), then in our minds we read $\operatorname{Prov}_{\text{PA}}(P)=:$ [PA $\vdash P$] as a kind of white box, that actually communicates the (true, genuine in the real world) existence of/"contains" a PA proof that concludes with $P$.

Like the subtle mistake is that people might interpret the Löb hypothesis [$\text{PA} \vdash (\operatorname{Prov}_{\text{PA}}(P) \to P)$] as: ([PA $\vdash \operatorname{Prov}_{\text{PA}}(P)$] $\implies$ [PA $\vdash P$]), i.e. "if PA says it proves $P$, then it does actually prove $P$". But this is implication in the "outer/meta" world, hence denoted by the "$\implies$" symbol! Only in the "outer/meta" world can we think of the precondition [PA $\vdash$ [PA $\vdash P$]] as a white box that actually has some meaning (semantics), as opposed to in the "inner" world where it's just a black box that is pure syntax.

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So the conclusion is that the funny Turing machine $N$ "wraps" $P$ in an extra "Prov" predicate, and we need the Löb hypothesis to "unwrap", to complete the full "double lap". And this gives us a proof of Löb's theorem: [PA $\vdash (\operatorname{Prov}_{\text{PA}}(P) \to P)$] $\implies$ [PA $\vdash P$].

(An "outer/meta" world implication "$\implies$". I'm not sure if is also an "inner" world implication "$\to$". If someone could clarify further in the comments, I would be grateful.)

(PLEASE please note I am just an amateur; I could be completely and utterly wrong about everything in this answer)