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[personal profile] gusl
Bringsjord has worked out my Penrose-like idea, formalizing his argument in quantified modal logic.

So the Halting Problem, HP becomes: forall TMs x, there exists a TM y such that x does not decide y.

He shows that the assumptions:

(1) I!: There exists a Turing Machine M, such that no Turing machine can decide whether it halts. And *necessarily* so, since it's a mathematical theorem. (this seems wrong to me!)
(2) For every Turing Machine M, there is a person S such that it's logically possible for S to decide M.
(3) All people are machines.

lead to a contradiction. (the contradiction is trivial, but he goes through the formal steps of his modal logic)


Of course, (2) is controversial. If I am a computer (which I believe I am), I would like to see my "Goedel sentence": which Turing Machine can I in principle not decide?

The lesson from Goedel's incompleteness theorem is that you always need to pick more axioms.

Analogously, if you're a Turing Machine whose mission in life is to decide whether Turing Machines halt (my new favorite way for thinking about this stuff, thanks to [livejournal.com profile] r6), you always need to change your Universal Turing Machine to one that decides more TMs.

Then the question becomes: "But how?". To me, the paradox remains, because if you have a systematic way of changing your Turing Machine simulator, then you're just using a meta-Turing Machine which is just as susceptible to undecidability: you'll never be able to decide whether the halting construction for meta-TM halts.

See Bringsjord - A Modal Disproof of "Strong" Artificial Intelligence (page 8)

(no subject)

Date: 2005-04-09 09:30 am (UTC)
From: [identity profile] gustavolacerda.livejournal.com
Well, in contrary to what I have written, I! is not the same as (1) (you'd have to look at the paper, page 8).

[livejournal.com profile] jcreed doesn't believe in (1), and I am also skeptical.

Apparently, there's mistake pushing the [] through the quantifiers and implications.

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