Re: a question
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sat, 29 Dec 2007 07:38:01 -0600
On Wed, 26 Dec 2007 10:24:28 -0800 (PST), aatu.koskensilta@xxxxxxxxx
wrote:
Gc wrote:
Let`s suppose a non-recursive extension of the Robinson arithmetic
"proves" every true arithmetical statement. Is this theory necessarily
complete?
Not necessarily, if the theory is in a language other than the
language of arithmetic. The theory ACA_0 + all arithmetical truths,
for example, is incomplete, as is ZFC + all arithmetical truths.
Possibly someone should point out that there are also utterly
trivial examples, not requiring any knowledge of anything
whatever:
Say L is the language of RA and T is the theory consisting of
all sentences of L that are true in the standard model.
So T is complete, or to be more specific, (L, T) is complete
(T is a complete theory in the language L.)
Now let L' be L with one constant symbol "c" added.
Let T' be all the consequences of T in the language
L. Then it's clear that (L', T') is incomplete
(since for example T + "c=0" and T + "~(c=0)"
both have a model.)
************************
David C. Ullrich
.
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