Newsgroups: comp.ai.philosophy
Path: cantaloupe.srv.cs.cmu.edu!das-news2.harvard.edu!news2.near.net!howland.reston.ans.net!pipex!uunet!psinntp!scylla!daryl
From: daryl@oracorp.com (Daryl McCullough)
Subject: Re: Godel, Lucas, Penrose, and Putnam
Message-ID: <1994Dec29.141029.1121@oracorp.com>
Organization: Odyssey Research Associates, Inc.
Date: Thu, 29 Dec 1994 14:10:29 GMT
Lines: 81

Lupton@luptonpj.demon.co.uk (Peter Lupton) writes:

>chalmers@bronze.ucs.indiana.edu (David Chalmers) writes:
>
>>Penrose's second argument proceeds, in essence, by saying "suppose I *am*
>>formal system F", and then arguing that with this assumption in place, he
>>can see the truth of statements that are beyond F's powers, even if F is
>>supplemented by the assumption that it is F.  Let F' be the system derived
>>by supplementing F with the assumption that it is F.  Then for the usual
>>reasons F' cannot see the truth of its Godel sentence G(F').  But Penrose
>>argues that (under the assumption that he is F) he can see the truth of
>>G(F'): he knows that he is consistent, so (under the assumption that he is
>>F) he knows that F is consistent, and indeed that F' is consistent, so he
>>knows that G(F') is true.  This contradicts the initial assumption that he
>>is F.  Similar reasoning shows that he cannot be any formal system.
>
>This argument by Penrose seems to be unsound and unsound in a
>very specific way. Clearly, F' is not the same formal system as F.
>That is plain enough, since the formal system F is not supposed
>to know that "I am F", but F' has, as an axiom, that "I am F".

Peter, I know we already hashed this over in email, but I would just
like to summarize for comp.ai.philosophy why I think you are wrong
about this.

You are right, that as Penrose described things, it doesn't make much
sense for F' to have as an axiom "I am F", which is false (for F').
However, it is easy to rephrase Penrose' argument so that it doesn't
go through this step.

Suppose for a second that Penrose is (unbeknownst to his closest
aquaintenances) actually a robot, and everything that Penrose could
ever accept as "unassailably true" is captured by a formal system F.
Now, among Penrose' unassailably true beliefs is the belief that his
reasoning powers are sound---that he never believes any false
statements. Since F captures all of Penrose' beliefs, there must be in
F a predicate symbol P(x) with the interpretation for Penrose that
"Penrose accepts that the statement whose Godel code is x is
unassailably true." Penrose' belief that he is himself sound implies
that for every statement A, F has a corresponding axiom

         P(#A) -> A

(using #A to mean the Godel code of A).

In other words, if Penrose believes A, then A must be true (since he
never believes any false statements.)

Now, since F is a computable theory, there will also be a definable
r.e. predicate F(x) meaning "x is the Godel code of a statement
provable in theory F." Penrose of course, may not know that F(x) and
P(x) are the same, but he can certainly reason that *if* F(x) is
the same as P(x), then F(x) must be sound. In other words:

         ((forall x F(x) <-> P(x)) & F(#A)) -> A

Now, consider a new theory F' which is formed by adding the axiom
(forall x F(x) <-> P(x)) to F. Working within F', we can see that it
is impossible for F to prove ~(forall x F(x) <-> P(x)) (the negation
of the new axiom), because F never proves false statements. From this
it follows that F + (forall x F(x) <-> P(x)) must be consistent.  (For
any statement A, if F + A is inconsistent, then F proves ~A.)
Therefore, working inside F', we can prove that the r.e. theory formed
by adding (forall x F(x) <-> P(x)) to F must be consistent.  In other
words, F' can prove that F' is consistent. But by Godel's theorem,
that is impossible, unless F' is actually inconsistent.

Therefore, we conclude that F' must be inconsistent. Since F' is
simply F + (forall x F(x) <-> P(x)), it follows that F proves 
~(forall x F(x) <-> P(x)). Therefore, F proves false things, and so F is
unsound. This contradicts either the assumption that Penrose'
reasoning is sound, or the assumption that Penrose' reasoning is
captured by F.

So if Penrose' reasoning is sound, and he believes it is sound, then
it is not computable.

Daryl McCullough
ORA Corp.
Ithaca, NY

