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From: George Greene on 6 Aug 2010 13:41
On Aug 6, 11:00 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> > > (2)  The more general matter that a formula with symbols that
> > > do not occur in a given set of axioms may still be derivable from that
> > > set of axioms.

I retorted:

> > NO, you do NOT get to derive Pv~P in a language that does not have P
> > in it.


> I didn't claim otherwise.

You lying ignoramus, you DID SO TOO claim otherwise.
Here is YOU CLAIMING it, AGAIN:

> > > a formula with symbols that
> > > do not occur in a given set of axioms may still be derivable from that
> > > set of axioms.
From: George Greene on 6 Aug 2010 13:43
On Aug 6, 11:00 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> The language may have '0' as a primitive symbol, but '0' might not be
> mentioned in the axioms. Still, if 'Ax x=x' is an axiom, then 0=0 is a
> theorem.

How exactly do you think 0 GOT INTO the language if not via an
axiom???
From: George Greene on 6 Aug 2010 13:47
On Aug 6, 11:00 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> The language may have '0' as a primitive symbol, but '0' might not be
> mentioned in the axioms.

That is debatable to some and ridiculous to me.

> Still, if 'Ax x=x' is an axiom, then 0=0 is a
> theorem. It's as simple as that

It IS NOT as simple as that! You have COMPLETELY MISUNDERSTOOD
INSTANTIATION in first-order semantics! It is THE STANDARD view AND
NOT my
contrarian one that the quantifier has to get instantiated to AN
ELEMENT OF THE DOMAIN
AND NOT to a constant of the language! MUCH DEPENDS on this!
What you MEANT to say was that the intended domain had (e.g.) 0 AS AN
ELEMENT,
but as an ANONYMOUS element, i.e., an element for which the language
had no name.
The (possible) existence of anonymous elements is key to G1 and a
whole bunch
of related results. But talking about parts OF THE LANGUAGE that are
not mentioned
in the axioms IS VACUOUS.
From: George Greene on 6 Aug 2010 13:50
> On Aug 6, 6:08 am, George Greene <gree...(a)email.unc.edu> wrote:
> > NO, you do NOT get to derive Pv~P in a language that does not have P
> > in it.

On Aug 6, 11:00 am, MoeBlee <jazzm...(a)hotmail.com> wrote:
> I didn't claim otherwise.

Pshaw.
> The language may have '0' as a primitive symbol, but '0' might not be
> mentioned in the axioms. Still, if 'Ax x=x' is an axiom, then 0=0 is a
> theorem. It's as simple as that

If THAT is simple, then it is JUST as simple to have an unnamed/unused
(by the axioms)
PROPOSITIONAL constant (which is which is what I had) or 0-ary
PREDICATE in the language, AS it was
to have an unnamed/unused (by the axioms) 0-ary TERM-functor (i.e. a
constant),
which is what you have. Neither is any more or less defensible than
the other,
and to insist on one while claiming not to be insisting on the other
is NOT supportable!
From: George Greene on 6 Aug 2010 13:51
On Aug 6, 10:40 am, stevendaryl3...(a)yahoo.com (Daryl McCullough)
wrote:
> I would say that logic today includes both deductive theories and models
> of languages.

You're attempting to debate something that is not under debate.
OF COURSE there is a model, and OF COURSE the purpose of
the deductive theory is to deduce truths within/about the model.
But the point is, the MODEL IS ONE thing, and the theory IS ANOTHER.
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