p j wrote:
> Thank you very much for your detailed explanation,
> y (~.,m)}~ (,m) *.//. >,x <@(<"1) m{y
>
> If I understand the parsing rule for [EMAIL PROTECTED]
> clip 'f@' right out of the sentence, parse it while
> consuming any .x dyadic arguments. then apply f to the
> result.
Yes.
And f is applied at the same rank as g.
Note also that this would have been equivalent:
y (~.,m)}~ (,m) *.//. >,x <@<"1 m{y
As would this (see Brian Schott's message):
y (~.,m)}~ (,m) *.//. ,/x <"1 m{y
> Are there any exceptions to this rule? - or clearer
> way to describe it?
The dictionary definition of @ seems pretty good, to
me.
> -- if there would be any trains that would be parsed
> after f is substituted back in, do they apply?
> ie
> h x [EMAIL PROTECTED] y = (h f) (x g y)?
First off, for general equivalence statements it's
better to work with @: than with @
@: doesn't impose the rank of g on the use of f
I'd have expressed your identity as
(h x f@:g y) -: (h f) (x g y)
since that better captures the niceties of J's formalisms.
And, yes, expressed this way (with the stipulation
that h, x, f, g and y are all verbs) you've got a tautology.
--
Raul
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