Re: Presburger arithmetic proves godels incompleteness theorem wrong
- From: David C. Ullrich <dullrich@xxxxxxxxxxx>
- Date: Sun, 23 Mar 2008 09:43:45 -0500
On Sun, 23 Mar 2008 00:07:59 -0500, "elsiemelsi"
<cyprinsam@xxxxxxxxxxxxxxx> wrote:
The australian philosopher colin leslie dean shows that Presburger
arithmetic proves godels incompleteness theorem wrong
Gödel's first incompletness proof is said to have showed that for
there any sufficiently powerful and consistent finitely axiomatizable
system there is a statement that cannot be shown to be true,ie systems
are undecidable
Gödel's second incompleteness theorem establishes that logical systems of
arithmetic can never contain a valid proof of their own consistency.
both these theorems are proven wrong by Presburger arithmetic
Presburger proved that his arithmatic was both consistent and decidable
thus disproving
godels theorems
http://nostalgia.wikipedia.org/wiki/Presburger_arithmetic
Presburger arithmetic is the first-order theory of the natural numbers
with addition. It is not as powerful as the Peano axioms because
multiplication is omitted. In fact, Presburger proved in 1929 that there
is an algorithm which decides for any given statement in Presburger
arithmetic whether it is true or not. No such algorithm exists for general
arithmetic as a consequence of the negative answer to the
Entscheidungsproblem. Furthermore, Presburger proved that his arithmetic
is consistent (does not contain contradictions) and complete (every
statement can either be proven or disproven). Again, this is false for
general arithmetic; this is the content of Gödel's incompleteness
theorem.
godels theorems can be applied to peano and robinsion arithmetic systems -
but not Presburger arithmetic
thus showing godels theorems are invalid
It shows that either Godel's theorems are wrong or
Presburger arithmetic does not satisfy the _hypotheses_
of Godel's theorems. Your statement above is conveniently
vague: exactly what does "sufficiently powerful" mean,
and exactly how do you know that Presburger arithmetic
_is_ sufficiently powerful in the required sense?
David C. Ullrich
.
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