Re: What is the intuitive meaning of Goedel's undecidable statement?
- From: "george" <greeneg@xxxxxxxxxx>
- Date: 13 Apr 2006 09:23:19 -0700
Bhupinder Singh Anand wrote:
Like all conjuring
tricks, Goedel's proof masks an extremely subtle misdirection. It
misleads even Goedel into presenting his assumption of the
omega-consistency of PA as an "intuitively unobjectionable" assumption.
That PA is omega-consistent is OF COURSE unobjectionable,
SINCE IT IS TRUE. Whether it is intuitively unobjectionable is
harder. It is provable from the compactness of first-order logic,
but that (arguably) wasn't "intuitive" either.
Once we accept this assumption as "intuitively unobjectionable", the
standard interpretations of Goedel's formal reasoning,
You need to be careful here: "interpretations" can have two meanings
in this technical context and you are not clarifying which way you
intend
"interpretations" to be interpreted here. One way (the one that
matters)
is about the standard interpretation or standard model of PA itself;
that is the natural numbers. The other way, you could mean the way
in which philosophers standardly interpret Godel's reasoning and its
implications.
and its consequences (which are based, essentially, on Goedel's informal
interpretations of his own formal reasoning in his seminal 1931 paper),
This is ridiculous. No formal anything NEEDS any "informal
interpretations",
ESPECIALLY at first order. Godel proved that FIRST, in the
COMPLETENESS
theorem. Anything you prove formally is provable precisely in virtue
of the
fact that it is INdependent of ALL interpretations; interpretations
NEVER
matter for what IS provable (at least not up to truth-value).
Interpretations
become relevant for sentences that are NOT provable.
the "logical conjuring trick" succeeds to baffle.
Although Rosser's proof avoids the assumption of omega-consistency, it,
in turn, masks an equally subtle (and, prima facie, also "intuitively
unobjectionable") misdirection.
It assumes that, from the assumption '[(Ex)R(x)] is PA-provable', we
may formally conclude, in a proof sequence, the existence of an '[s]'
such that '[R(s)] is PA-provable', and arrive at a contradiction.
That is absolutely indisputably true. It is provable.
If Ex[Rx] is provable, then by the completeness theorem,
it must be true in EVERY model, and THEREFORE it must
be true IN THE STANDARD model. In the standard model,
every number is named by a numeral, so that numeral MUST BE
that '[s]'.
It, then, concludes that we cannot have '[(Ex)R(x)]' as a PA theorem.
Well, obviously, you can't, if PA is consistent. Not
"omega-consistent",
or 1-consistent, but JUST PLAIN consistent, PERIOD. I.e., not if PA
has ANY MODELS AT ALL.
Now the misdirection underlying Goedel's reasoning only becomes
apparent when we note that, post-Goedel - the standard Tarskian
definition of the truth of a formal expression, say [R(x)], of a
language L, under an interpretation M, is defined in terms of the
'satisfiability' of the corresponding interpretation of the expression
in M, say R(x).
No, it isn't. It's defined in terms of the satisfACTION of R(x) under
M.
It is completely ignorant and illiterate to speak as though R(x) could
be satisfiABLE under M. Once you choose M, there is no more modality,
there is no more possibility or variation. M by definition SIMPLY
SETTLES
THE QUESTION of the truth-value of R(x), for all R. SatisfiABILITY
relates
to whether there could exist ANY M that makes R(x) true. Once you fix
the M, R(x) is either true or it isn't. Its truth under ANY ONE M
implies
satisfiABILITY *globally*, ALL the time. M has an opinion about
whether
R(x) is satisfIED, NOT about whether it is satisfiABLE (unless M's
opinion
is that R(x) is satisfied, in which case that implies that it is
satisfiable;
but if M opines that R(x) is not satisfiED, that does NOT imply that
R(x)
is unsatisfiABLE).
However, the definition is silent on how, or even whether, such
'satisfiability' is to be effectively determined in any particular
case.
It doesn't matter. If M in fact does determine a truth value then
everything is fine. It doesn't matter whether the determination is
effective or not.
(The seeds of Platonic interpretations of Goedel's reasoning -
including his own - lie in the acceptance that such 'satisfiability'
need not be effectively determinable in every case.)
Well, fundamentally, IT CAN'T be.
Take N as the primary obvious logical natural example.
Satisfiability of universal generalizations in N is not effectively
confirmable (equivalently, unsatisfiability of existential
generalizations
is not either). If you try ANY other tack (i.e., if you try to require
that all of a model's answers to all truth-value questions be
"effectively"
determinable) then you essentially require all models to be
recursive/decidable
sets. Since N ITSELF (the smallest POSSIBLE) model, has non-recursive
subsets, this is just obviously a ridiculous AND UNSATISFIABLE
requirement.
.
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