Re: Simple yet Profound Metatheorem

sradhakr says...

>I really doubt whether you can claim that anything is proven in a
>*meaningful* sense without the use of the law of non-contradiction. For
>example, suppose I make the following assertions, given your proof of
>the above schema:
>
>(1) P
>
>(2) P -> Q
>
>(3) ~Q
>
>Now you can object that the conclusion (3) is not acceptable because it
>contradicts the given schema.

You can make whatever assertions you like. The question is
a matter of proof. A random collection of assertions is not
a proof.

>But you need the law of non-contradiction
>in order to make the said objection.

The law of contradiction isn't relevant to the question of whether
(1), (2), (3) is a valid proof.

>The "proof" may go through formally, but your argument for the
>law of non-contradiction below actually requires that "False
>cannot be the case" in a meaningful sense.

No, that's irrelevant to the proof of ~(P&~P).

>> Then a special case is
>>
>> (P & (P -> False)) -> False
>>
>> which is the same thing as
>>
>> ~(P&~P)
>
>To arrive at this conclusion a reductio is tacitly presumed.

No, it's not.

>When you interpret ~P as P->False, what you mean is that "Any
>proof of P can be converted to a proof of False", which in turn
>yields "A proof of P is impossible"

No, a proof of ~P is *not* a proof that "A proof of P is impossible".
If you have inconsistent axioms, then you can prove both P and
also ~P.

A proof of ~P amounts to a proof that "P implies anything". So
if *anything* is not provable, then P is not provable. But it's
possible that your axioms allow *everything* to be proved.

>The problem, I claim, is that you need a prior conception that "False"
>cannot be the case

Nothing in my proof depends on the assumption that False cannot
be the case.

>I cannot conceive of any way in which you can *prove*
>False without the use of the law of non-contradiction.

Hopefully, you can't prove False, so it is *good* that
you cannot conceive of any way to prove it.

>But how is it justified from the intuitionistic
>standpoint?

Intuitionistic proof doesn't require the assumption that
False cannot be the case, it only requires the assumption
that "False implies anything", which is true by definition
of "False".

>Remember that intuitionism requires all truths to be stated
>with proof and we are trying to prove the law of non-contradiction
>here.

The law of non-contradiction is just

(P & (P -> False)) -> False

>If we don't have any prior conception that False cannot be the
>case, the argument given in my previous post shows that you can
>interpret the given proof in two ways -- one favourable and the other
>unfavourable.

I couldn't make any sense of that post. Whether ~(P&~P) is provable
or not does not depend on "interpretations", it depends on your
rules of inference and your axioms.

>When we choose to make the favourable interpretation, we
>are tacitly asserting that we wish the law of non-contradiction to be
>true, and this is nothing more than an axiomatic assertion, without
>proof. It seems to me that this business of trying to "prove" the law
>of non-contradiction in a non-circular, meaningful sense is doomed from
>the start.

Well, I have no idea what you are talking about. I gave you
a proof of ~(P&~P), and that proof was noncircular. So what
you are saying doesn't make any sense to me.

I think you are talking about *interpretations* rather than
*proof*. There is no circularity in the proof of ~(P&~P). It
doesn't rely in any way on the notion that "There is no
proof of False", so your complaints about it are just silly.

Now, when it comes to *interpretations*, you can legitimately
ask: "How do we know that it is never the case that P&~P?"
Well, what we can say is that if, for any statement Q,
it is not the case that Q, then in particular it is not
the case that P&~P. If *anything* is not true, then False
is not true. That's the best you can do.

--
Daryl McCullough
Ithaca, NY

.

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