Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable
- From: "george" <greeneg@xxxxxxxxxx>
- Date: 18 Nov 2006 20:52:54 -0800
Newberry wrote:
If
(P & ~P) --> Q (1)
ought not to be derivable then Goedel's formula ought not to be
derivable either.
This is just ridiculous, but without even getting into all
you don't know about Godel's theorem, we need to begin
by concentrating on all you don't know about indirect proofs.
We must begin by stressing that NOBODY EVER uses
(P ^ ~P) -> Q to infer Q as a truth, as a final conclusion.
It is obviously worthless for that, because it could just
as easily be used to infer ~Q. This is always an intermediate
step in a proof and the real goal is ALWAYS something else.
Nobody intentionally starts with a contradiction.
A contradiction is a subgoal along the way to proving something
more important, namely, that one of the original premises was
incorrect. Basically, if you need one PARTICULAR contradiction,
OTHER than P ^ ~P, as a piece to match a pattern for another
inference rule, then this rule allows you to trade P ^ ~P for the
one you need. In particular, if, having assumed R and P, you can
derive ~P or ~R, then you are entitled to conclude ~R and ~P.
NOT from
(P ^ ~P) ->Q (with R as Q), but just (as DMC has already explained)
as a conference of inference rules DEFINING "~".
Not that ~ *needs* to be defined with inference rules;
~ is just a unary boolean function; there are only 2 of these;
~ is the one that is not the identity. Equivalently, it is the non-
identity permutation of 2 elements.
.
- References:
- Prev by Date: Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable
- Next by Date: Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable
- Previous by thread: Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable
- Next by thread: Our Heavenly Father on 11-22-05 at 12:47 P.M.
- Index(es):
Relevant Pages
|
