--- Roger Hui <[EMAIL PROTECTED]> wrote:
> Since APL days u/y is {.y when 1=#y , even when
> y is not in the domain of u . For example:
>
> +/'a'
> a
>
> It is not much of a flaw. The only plausible
> alternative is to signal error.
>
> The identity that is used to reason about u/y is
>
> u/y <-> (u/k{.y) u (u/k}.y)
>
> for k e. i.1+#y .
>
k=. 0
u=. ,
y=. 1
#$(u/k{.y) u (u/k}.y)
1
#$u/y
0
k=. 1
#$(u/k{.y) u (u/k}.y)
1
#$u/y
0
(u/y) -: (u/k{.y) u (u/k}.y)
0
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