From: daryl@oracorp.com (Daryl McCullough)
Subject: Re: Human intelligence vs. Machine intelligence
Organization: ORA Corporation
Date: Thu, 15 Oct 1992 17:16:36 GMT

In article <SMAILL.92Oct11171415@affric.aisb.ed.ac.uk>,
smaill@aisb.ed.ac.uk (Alan Smaill) writes:

>   That much is true. Where Penrose jumps off the deep end is in
>   concluding that therefore no theory can formalize human reasoning. The
>   most that can be concluded is that *if* there is a theory that
>   formalizes all of human reasoning, then it must be a theory so complex
>   that we can't know whether it is consistent or not.
>
>Lest anyone thinks that Penrose is unaware of this step in the argument,
>this is what he says: (p418, hardback edition (alright, softback too),
>arguing on the assumption that a "universal" theory exists)
>
>"Thus, we are driven to the conclusion that the algorithm that
>mathematicians actually use to decide mathematical truth is so complicated
>or obscure that its very validity can never be known to us.
>  But this flies in the face of what mathematics is all about! ...
>Mathematical truth is not a horrendously complicated dogma whose validity
>is beyond our comprehension.  It is something built up from such simple
>and obvious ingredients -- and when we comprehend them, their truth
>is clear and agreed by all.
>  To my thinking, this is as blatant a _reductio ad absurdum_ as we
>can hope to achieve, short of mathematical proof."

The crucial part of the argument hinges on the question of whether
humans can recognize arbitrary valid mathematical statements. If we
can, then Penrose is right that we can do something that no machine
can do. However, what kind of argument does Penrose advance that
humans are capable of this? A ridiculous one, in my opinion. Just
because mathematical theories are built up from "simple and obvious
ingredients" does *not* imply that the consequences are simple and
obvious. For example, nobody has the faintest idea whether Fermat's
Last Theorem follows from the simple and obvious axioms of PA (Peano's
axioms for arithmetic). Therefore, the theory PA + FLT *might* be a
valid theory, and it might not be. When I say that a theory might be
too complicated for us to recognize whether it is consistent, I don't
mean anything any more complicated than PA + FLT. The axioms are not
complicated, but the consequences of the axioms are.

>   For example, let's examine whether it is possible for all human
>   reasoning about arithmetic to be formalized in the theory NF (Quine's
>   New Foundations, a variant of set theory which is different from ZFC
>   but is not known to be consistent relative to ZFC). We can do Penrose'
>   trick of coming up with a sentence G such that *if* NF is consistent,
>   then (1) G is true, and (2) NF doesn't prove G. Do we know that G is
>   true? No, because G is only true if NF is consistent, and we don't
>   know whether NF is consistent. Therefore, Penrose' trick fails to come
>   up with a sentence which we know to be true and which NF can't prove.
>   In other words, Penrose is mistaken.
>
>I'm not convinced by this example.  
>
>What if NF is shown to be inconsistent?  Then it is useless as a basis
>for an artificial reasoner (where I make the reasonable assumption
>that any consequence of NF is attainable by the artificial reasoner).
>So, where the trick doesn't work, the system is not justifed as a basis
>for an artificial reasoner either.

Well, NF *hasn't* been shown to be inconsistent. Therefore, if Penrose'
argument is valid, then it should not depend on whether NF is consistent
or not.

You are assuming that there are only two possibilities: (a) NF is
consistent and we can prove that it is consistent, or (b) NF is
inconsistent.  There is a third possibility: (c) NF is consistent, but
we can't *prove* that it is consistent. In cases (a) or (b), it does
indeed follow that NF is inadequate for formalizing all of human
mathematical reasoning. However, in case (c), there is no
contradiction in assuming that NF captures all of human mathematical
reasoning. For Penrose' trick to work, it is not enough that a theory
be consistent--he must also *know* that it is consistent.  He must be
able to rule out case (c), not only for NF, but also for *every*
theory more powerful than Peano Arithmetic.

There is nothing in any of Penrose' arguments that rules out the
possibility of a theory T such that:

    1. T is consistent.
    2. T is more powerful than Peano Arithmetic.
    3. T captures all of human reasoning about arithmetic.
    4. It is impossible for human reasoning to prove that T is consistent.

The mistake Penrose makes is in assuming that because something is
true, then there exists a proof (or a convincing argument) that it is
true. A theory T can be consistent without our being able to give a good
argument for why it is consistent.

Daryl McCullough
ORA Corp.
Ithaca, NY