Re: Penrose vs the Robot

Rupert says...

>What about Penrose's second argument?
>
>Penrose doesn't know whether he is F, but let F' be F supplemented by
>the assumption that Penrose is F. On the assumption that he is F,
>Penrose can see the truth of the Goedel sentence for F'. Therefore, if
>Penrose is actually F, F' proves its own Goedel sentence and therefore
>is not sound.

I've been thinking about it, and here's how I think it's possible
to have the following:

1. Penrose' beliefs about arithmetic (or at least pi-1 sentences)
are sound.

2. Penrose believes himself to be consistent.

3. Penrose' set of beliefs is r.e.

Let T_0 be some r.e. theory extending PA. Create a new theory, T
by adding two new predicate symbols B(x) and M(x) and some new
rules of inference. The interpretatios of B and M are:

B(x) == Penrose believes the sentence whose code is x, and
M(x) == Penrose considers the sentence whose code is x to be meaningful.

The rules of inference for T are the following:

1. If T_0 |- A then T |- A
2. If T |- A -> C, and T |- A, then T |- C
3. If T |- A and T |- M(A) then T |- B(A)
4. T |- B(A) -> B(B(A))
5. T |- B(A -> C) -> (B(A) -> B(C))
6. T |- B(A) -> A

(where B(A) and M(A) really mean B(n) and M(n)
where n is the Godel code of A).

Now, it seems to me that the usual Godelian/Tarskian trick
to get a contradiction is blocked: Form G such that

T |- G <-> not B(G)

Then we can go through the usual steps:

7. G <-> not B(G)
8. B(G) -> B(B(G)) (from 4)
9. B(G) -> B(not B(G)) (from 7)
10. B(G) -> B(B(G) and not B(G)) (from 8 and 9)
11. B(G) -> B(false) (from 10)
12. B(G) -> false (from 6 and 11)
13. not B(G) (from 12)
14. G (from 7 and 13)

So G is a theorem of T. However, we *don't* get to conclude
B(G) because the rule 3 requires that G be meaningful. As
a matter of fact, T can prove that G is *not* meaningful.
I think it is a consequence of this theory that

15. not M(G)

This set of axioms is clearly consistent, because
the interpretation B(x) = M(x) = false (Penrose
doesn't believe anything, and doesn't find anything
meaningful) satisfies all the axioms. The question
is whether there exists a nontrivial interpretation
of B(x) and M(x). In particular, I'd like to add
the axiom:

If A is arithmetical, then T |- M(A)

In that case, I think T is still consistent, but
it becomes inconsistent if you add the axiom

T |- B(x) <-> Th(x)

where Th(x) expresses theoremhood in some r.e. theory.

--
Daryl McCullough
Ithaca, NY

.

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