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Fephisto wrote:
> herbzet wrote:
[...]
> > If P ^ Q = R then you're going to need some inference rule to
> > formally deduce R from the axioms P and Q. Even if R is
> > literally the formula "P ^ Q" it still needs to be formally
> > deduced from the axioms (unless it already _is_ an axiom,
> > which you have not said).
> >
>
> Alright, I've looked up meta-variables and propositional variables on
> Wikipedia, but both send me to another page to another page to another
> page....etc.. Do you at least have some good sources on this stuff?
Yes. "Formal Logic" by A.N. Prior [1962] is an awesome introductory
book. There is no one who writes on logic with the clarity of Prior.
If your library doesn't have it, ask them if they are part of a library
loan program. Worth the trouble.
Also, if I remember correctly,
1) "Principia Mathematica" by Russell & Whitehead
2) "Principles of Mathematical Logic" [1950] by Hilbert & Ackermann
3) "Introduction to Logic" by Copi. My copy is the 4th edition [1972].
Quine's "Mathematical Logic" revised [1951] is a good book, but
he doesn't use propositional variables at all, just meta-variables
phi, psi, chi, etc.
I looked for references on the Web. Most, but not all, that I
checked use upper-case P, Q, R, ..., for propositional variables,
whereas the books I referred to use lower-case (if I recall correctly).
Try googling "propositional calculus".
Apparently, a sea-change has occured in notation since the books
I mentioned were published.
> > > 2) As for rules I'd assume I'd being using the three classical laws
> > > (excluded middle, contradiction, and bivalence), and I should be able
> > > to do my ordinary implications, contrapositives, and reductio ad
> > > absurdums.
> >
> > I take this as meaning that you are introducing your two axioms
> > P and P -> Q (or in system 2, your two axioms P and Q) as being
> > added to the normal set of axioms.
> >
> > Could you clear up this point please? Are you adding your axioms
> > to a pre-existing system? Is that system the propositional
> > calculus (also called the sentential calculus)? Or is it the
> > predicate calculus? Or what?
> >
>
> Whatever logic system set theory is based off of, I'm using this,
> because the statements that I've written are all written using set
> theory definitions, and until now, that's the only logical system that,
> of, I was aware.
OK, that would probably be first-order predicate logic extended with
ZFC set-theoretic axioms.
> > > > In the present
> > > > cases, it doesn't matter, if MP is your sole inference
> > > > rule, because the same considerations apply to ~R.
> > > >
> > > > I don't see any necessity, at this point, to be considering
> > > > the systems formed by negating the axioms. Is there a
> > > > reason to do so?
> > > >
> > > > Am I making sense to you?
> > >
> > > If I have R, and two axioms P and Q,
> >
> > I don't know what you mean by "If I have R", although I understand
> > that we have the two axioms P and Q. Possibly we can just ignore
> > this introductory clause you have written without loss of meaning?
> >
>
> I meant R just to be some other random statement in (clarified just a
> little bit ago I guess) in set theory.
>
> > > and I happen to prove that ~P ->
> > > R, then R is independent from R
> >
> > I take it that you mean "then R is independent from P".
> >
> > (that is, as I'm taking it P cannot
> > > imply R), because if R was dependent then P ^ ~P -> R, which would be a
> > > contradiction,
> >
> > I'm not following you here. You seem to be saying that if ~P implies
> > R, then (if the system is consistent) it is not possible that P should
> > imply R. Is this what you mean?
> >
>
> Yes, by the law of contradiction. If P implies R, but then ~P implies
> R, then
Then R is true. Formally,
(a) ((P -> R) ^ (~P -> R)) -> R
is a tautology.
> it follows that both P ^ ~P -> R,
(b) (P ^ ~P) -> R
is also a tautology, whether or not P implies R, and
whether or not ~P implies R.
> but that would go against the
> law of contradiction,
You know, I think I've forgotten what the law of contradiction is.
Could you please state it, as rigorously as possible? Thanx.
> therefore either P implies R or ~P implies R.
> Thus if I prove that P implies R, I know that R is independent from ~P.
The usual usage is: If P implies neither R nor ~R, then R is independent
from P (and so is ~R)
Or actually, where S is a set of formulae, and phi is a formula:
(c) phi is independent of S iff ~(S |- phi) and ~(S |- ~phi).
Capeesh?
You seem to be under the impression that if P implies R, then ~P cannot
imply R without contradiction (or conversly, that if ~P implies R,
then P cannot imply imply R without contradiction.)
This is wrong; if R is true, it is quite possible that it will be
implied by both P and ~P. (See tautology (a) above)
You also seem to be under the impression (I'm less sure about this)
that R is independent of P if ~P implies R.
That is, you seem to think that:
(d) phi is independent of S iff (S' |- phi),
(where S' is the set of negated formulae of S).
This also is wrong. (See (c) above).
Examine tautologies (a) and (b) above. Since they are both tautologies,
they are therefore equivalent in truth value, but they have different
meanings.
Also, (c) is at least pretty close to the correct definition of
independence, but (d) isn't.
--
hz
'Even the crows on the roofs caw about the nature of conditionals.'
-- Callimachus --
--
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