Re: If (P & ~P) -> Q is not derivable then Goedel's formula is not derivable


Newberry wrote:
If

(P & ~P) --> Q (1)

ought not to be derivable then Goedel's formula ought not to be
derivable either. Here is why:


(1) is counter-intuitive and it ought not to be derivable. But it is
equivalent to

~((P & ~P) & ~Q) (2)

So it ought not to be derivable either. Analogically

~(Ey)((Py & ~Py) & ~Qy) (3)

ought not to be derivable. We note that

~(Ey)((Py & ~Py) (4)

Here you have simply *asserted* ~(Ey)(Py & ~Py). If you modify this to
~(Ey)(Py & ~Py) is *derivable* (or provable), I claim that is false. If
(1) is not derivable, then (4) is also not derivable. If you simply
assert (4) as you do in your argument, then you are led to the logical
fallacy below:


Therefore whenever

~(Ey)Fy (5)

then

~(Ey)(Fy & Gy) (6)

ought not to be derivable.


This is not correct, as pointed out by Peter Smith. The hypothesis (5)
leads to the conclusion (6).

In the logic NAFL (see <http://arxiv.org/abs/math.LO/0506475>), the law
of non-contradiction ~(P&~P) is not a theorem of a theory T in which P
is undecidable (i..e, neither P nor ~P is provable in T). This entails
that there must exist a model for T in which P&~P is the case. But
~(P&~P) is simply *asserted* in what I call the "proof syntax" of T,
i.e., to prove the theorems of T you *assume* ~(P&~P) (and the
classical inference rules). In other words, ~(P&~P) (and the classical
inference rules imply the theorems of T, but the theorems of T do not
imply ~(P&~P). Ex falso quodlibet fails in NAFL, i.e., you cannot
deduce an arbitrary proposition Q from the assumptions P and ~P. There
are more subtleties here that I will not go into for now.

Regards, RS

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