On 12 April 2011 01:00, Raul Miller <[email protected]> wrote:
> I cannot imagine any interpretations which conflict with my statement
> which do not also conflict with the dictionary entry.
>
> But perhaps you can show me one?

You said:

> Unless 0=#y (which is explicitly treated), u/y necessarily has _1+#y
> instances of u.  This follows immediately from the first sentence at
> http://www.jsoftware.com/help/dictionary/d420.htm

and the sentence you are referring to is
    `u/y applies the dyad u between the items of y'.

If you choose to maintain that, somehow,
    `0 applications between 1 elements'
has definite meaning, then it seems to me that, to be self-consistent,
you should go all the way through and also consider
    `-1 applications between 0 elements'
(in which case you would also have to explain how -1 applications
of a verb yield that verb's neutral element).

I prefer to stick by common sense and consider the DoJ statement
as applying only to 1<#y, since only then the word `between' is
meaningful.  Thus, both 1=#y and 0=#y require a separate explanation.


On 12 April 2011 01:15, Brian Schott <[email protected]> wrote:
> Actually, then (u/k{.y) u (u/k}.y) produces the correct domain error
> when y=:'!', doesn't it?

Yes, it is correct to produce an error.  But this error invalidates the
identity u/y ↔ (u/k{.y) u (u/k}.y), because its l.h.s. produces a normal
value.  And the identity does not generally hold for several other
reasons as well.

What I mean is that it is not correct to use a generally false identity
as the above, in order to infer the value (meaning) of u/y for 1=#y
(as I believe that was what you intended -- or am I mistaken?).
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