Date: Mon, 20 Oct 2008 20:09:51 -0700
From: Thomas Lord <[EMAIL PROTECTED]>
Cc: [email protected]
Alan Bawden wrote:
> Stop being so argumentative
>
I'm sorry I made a dumb mistake in my example but please
don't attack my character as a result.
Actually, it was a very interesting mistake. I thought the way in which
your proposed transformation failed was very enlightening. My cranky
reaction was not because your original mistake was "dumb", but because when
I pointed out the error, you continued to argue without realizing that I
had in fact pointed out an error.
Here's a better, simpler, corrected example:
(apply = x) ; true when the elements of x
; are all equal
<=>
(let ((xv x))
; the elements of x are all =
(not ; if it is false that
(exists ; there exists
(lambda (x0) ; some element of x ("x0")
(exists ; such that for that x0 there exists
(lambda (x1) ; another element of x ("x1")
(not ; where it is not the case that
(= x0 x1))) ; those elements are equal
xv))
xv)))
Now, to be clear, I think that second transformation is
also a bogus transformation because my opinion is that
(apply = '(q)) ought to be an error, not #t -- but if we stipulate
that "x" denotes a list of numbers then the transformation
is not bogus.
For `=' it is indeed not bogus. Unfortunately, substituting `<' for `=' in
the above equivalence results in something that -is- bogus. So while you
may be able to use this example to argue that `(=)' should return #T, you
can not use it to argue that `(<)' should return #T.
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