Re: The Law of the Excluded Middle again (long)

On Mon, 03 Dec 2007 23:35:30 -0500, quasi <quasi@xxxxxxxx> wrote:

On Tue, 04 Dec 2007 04:13:40 +0000, Angus Rodgers
<twirlip@xxxxxxxxxxx> wrote:

Recall the context of the argument. When the LEM is applied, as in
"(x >= 0) or not (x >= 0)", /if/ there is to be an objection on the
grounds that truth means provability, then the truth of a statement
certainly cannot be identified with its provability using itself as
a hypothesis, because such an identification would make every state-
ment true.

You lost me.

How in the world would it make any statement true?

When you make an assumption within a proof, anything you prove based
on that assumption is conditional on that assumption (together with
the original hypothesis, and any other enclosing assumptions in
force).

Thus, if you assume x >= 0 within a proof, then within the scope of
that assumption, x >= 0 is true, and provably so (but it would be
silly to prove it, since you already have it as a temporary
hypothesis). Still, if challenged, the proof is trivial ...

x >= 0 (reason: temporary hypothesis)

Therefore x >= 0 (reason: prior line)

You appear to be claiming that allowing such assumptions would make
every statement true. But that's silly. You can only prove those
things that are provable from the hypotheses (including any temporary
ones in force), and any conclusions obtained are conditional on those
hypotheses. Thus, I don't get your point.

I found it confusing to think about, myself, and it's possible that
what I wrote was gibberish, but even in the cold light of the morning
I still can't actually see anything wrong with it. (Apart, of course,
from the fact that my entire argument about "variables" has long ago
been shown to be irrelevant!) Of course absolutely every statement is
provable from itself as a hypothesis. Your entire argument rested on
the existence of such a trivial proof for every statement. You said
this refuted my assertion that there was no sense in which a statement
like 'x >= 0' could be said to be "provable". And you were right.
But my reply is that there is still no sense in which such a statement
could be said to be "provable" /and/ its truth could be identified
with its provability. (To be clear: what I am attempting to do here
is to rehash - at your request - an argument which has already been
refuted. Keith explained simply that my argument was misconceived;
I no longer have much interest in it. Nevertheless you keep asking
me to explain what on Earth I thought I meant by it, and I am doing
my best to oblige.) Remember the context of the argument: I thought
I had identified a reason, in this special case (the LEM applied to
propositions containing free variables and no quantifiers - I'm not
trying to be completely precise about this, because there's no longer
any point), why the LEM could not be objected to on the grounds (seen
in the other thread I started, a month or so ago, entitled "A quote
(and question) about intuitionism") that truth in mathematics means
provability. But (to repeat one last time) I was wrong: my argument
was a red herring, because as Keith pointed out, the /hypothetical/
truth of such a proposition as "x >= 0" /can/ be identified with its
/hypothetical/ provability (that is, by someone who believes in the
identification generally), without any special consideration having
to be made. (I hope I'm not misrepresenting Keith's argument - it's
possible, because I still don't at all understand his position in
general.)

It's convoluted; and as I keep saying, it's now pointless to analyse
it in so much detail; but you do keep pressing me, and for the moment
I think this is the best explanation I can give.
--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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