FOM: Truth of G
Raatikainen Panu A K
Praatikainen at elo.helsinki.fi
Tue Nov 7 09:14:20 EST 2000
There is much unclarity both in the literature and even in certain
recent postings in FOM on the reasons for holding the Goedel
sentence G (for, say, PA) true. Many think that one can intuitively
see that it is true because "it says that it is unprovable". Others
assume that we somehow check that it holds in the standard
model of arithmetic. Both these views are problematic and
irrelevant. But the true reasons seem to be unclear even to persons
well educated in mathematical logic. Let me try to clarify the issue.
The structure of Goedel's proof is (very roughly) the following:
ASSUME THAT (e.g.) PA is consistent (otherwise it proves
EVERY sentence and is trivially complete).
By diagonalization, one can construct a sentence G that is
independent (neither provable nor refutable) of PA.
So far so good, but how, then, can one conclude that G is true ?
There are even two somewhat different ways...
First, assuming that the provability predicate used is normal, one
can show (and prove even inside PA) that
G <-> Cons(PA).
(Although one can prove neither side of the equivalence in PA).
Therefore, the truth of the sentence is, in a sense, already
assumed in the beginning of the proof.
NOTE: No non-mechanical intuition, no use of the standard model,
The proof goes through also for a theory containing arithmetic that
is in fact inconsistent (e.g. Quine's original system). In that case,
the sentence is actually false, whatever one's intuitions are ...
BUT SECOND, there is also another, more general way to
conclude that an unprovable sentence is true. Let T be a theory
(assumed to be consistent) that contains Robinson arithmetic Q.
Let then A be an arbitrary Pi-0-1 sentence that has been proved, by
whatever method (model-theoretical, proof-theoretical, by utilizing
some paradox, etc.) to be independent of T. A need not have any
meta-mathematical "second meaning" - indeed, we need not
understand at all what it says. Still, we can conclude immediately
that A is true. WHY?
Because every theory extending Q is complete for Sigma-0-1
sentences (i.e. proves every true Sigma-0-1 sentence). Therefore, if
A were false, its negation would be true and, being Sigma-0-1,
provable (and A thus refutable). But A was shown to be
independent of T. Therefore it must be true.
Again, no intuition, no comparison to the standard model, is
I hope these little remarks help to demystify Goedel's sentence.
Department of Philosophy
University of Helsinki
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