On 11 April 2011 23:43, Brian Schott <[email protected]> wrote:
> Boyko,
>
> The expression I cited applies to "... k e. 0,#y .", of which 1
> is a possible value of #y. Isn't that good enough?
If you mean that (u/k{.y) u (u/k}.y) is equivalent to u/y, and that
that equivalence could be used to define u/y for 1=#y, then note
that the said `equivalence' does not hold for many, many values
of u, y, and k.
In your example, consider y=.'!' (keeping k=.1 and u=.-).
Then u/y appears to be '!' (so says the J interpreter), but
(u/k{.y) u (u/k}.y) is obviously erroneous (being equivalent
to (-/'!') - 0).
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