[FOM] Re: Reflection and the Lucas-Penrose Argument
Timothy Y. Chow
tchow at alum.mit.edu
Fri Sep 3 14:17:15 EDT 2004
On Thu, 2 Sep 2004, Jeffrey Ketland wrote:
> But it seems to me that the principle you've given is much too weak for
> non-algorithmicity. The principle
> (Ref) If S is acceptable then Con(S) is too
> is some sort of reflective closure condition. I agree with it, of course.
> Relatedly, any theory of truth worth its salt will prove
> If a set S of sentences is true, then Con(S) is true.
> But how does reflective closure imply that "the unassailably reliable
> truths can't be recursively axiomatizable"?
I tried sketching out this argument back in May, and missed a couple of
subtleties. Let me try it again.
Let S be a set of sentences in the first-order language of arithmetic.
We want S to somehow capture the arithmetic truths that are unassailably
knowable by humans. So it is plausible to assume that S contains the
axioms of PA and that S is consistent.
Now suppose that S is recursively axiomatizable. There's a subtlety
here, because many different recursive axiomatizations of S are possible.
Even if we take S to be something familiar like PA, someone could describe
a Turing machine M with the property that M "generates" PA iff (say) the
existence of measurable cardinals is consistent with ZFC (or any other
arithmetical sentence that happens to be true but that we don't know to
be true). Then M would furnish a recursive axiomatization of PA but we
might have trouble recognizing it as such.
This provides a loophole of sorts in the Lucas-Penrose argument; maybe
S is recursively axiomatizable, but any explicitly given Turing machine
describing S will be controversial---we won't be able to look at it,
understand exactly which statements are in question, and agree that
they're all unassailably true. However, in such a case I think the
Lucas-Penrose side would have lost a battle but won the war. One could
go around claiming that S is recursively axiomatizable but one would
not be able to back up such a claim by exhibiting a Turing machine and
letting people inspect it and convince themselves. This would not be
So let's turn to the interesting case, where we assume that it's actually
possible to exhibit a Turing machine M and become unassailably certain
that all the sentences M is generating---i.e., S---are true. Then we
can apply the usual construction to produce a sentence Con(S). According
to your reflective closure condition, Con(S) is also unassailably true.
So by definition of S, Con(S) is in S. But this contradicts Goedel's
2nd theorem. End of proof.
Your description of Feferman's reflective closure and of theories of truth
and so forth fail to get around this problem. In fact you hinted at this
yourself when you said you didn't understand why Feferman didn't iterate
the Ref operator. Indeed, if you find the reflective closure of PA to be
unassailably true and are able to write down an axiomatization of it and
definitively recognize it as such, then you can form the sentence
Con(Ref(PA)), and find it to be unassailably true.
Either S is not recursively axiomatizable, or it is but you can't write S
down explicitly and point to it and say for sure that it is S. Or, of
course, your reflective closure principle is wrong.
As I said before, Torkel Franzen has an excellent discussion of all this
in his book on inexhaustibility.
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