It's a theorem:
[x] >@(f each) y
[x] >@(f&.>) y
[x] >@((<@:f)&>) y
[x] (>@(<@:f)&> y
[x] (>@:<)@:f&> y
[x] f&> y
[x] (f every) y
Some of these steps are very much not obvious IMO. And you have to get
the rank of each right, that is, use the NuVoc definition of &. rather
than the Dictionary one.
Henry Rich
On 6/15/2018 8:30 PM, Ian Clark wrote:
I've checked Chapter 1 off, but that's only to say I've checked out the
code and verified it gives the results claimed. I didn't see it as my job
to rewrite the treatment to make it clearer – which I can't do anyway
without being sure what the author is trying to convey.
I must confess that first section completely baffles me. I cannot see how
to relate the "general rule" to actual examples of J code, although the
article goes on to do just that … it seems. Does the "rule" represent real
working J code? – even in a generic sense? Is it even true? (Theorems have
to be true, but rules only have to be obeyed.) If it isn't always true, am
I to understand it as a rule-of-thumb?And if it is in fact universally
true, what procedure must I, the novice reader, follow in order to convert
the "generics" into "specifics" to verify the fact?
I'd be grateful for someone to cast light on the matter. Failing which,
maybe I ought to remove my green checkmark, stand aside to let someone else
scratch their head over it.
On Sat, Jun 16, 2018 at 12:41 AM, David Lambert <b49p23t...@gmail.com>
wrote:
50 Shades of j chapter 1 now says that rule is completely general. I'm
somewhat weak on j transformations and proofs, although what was there was
incorrect because of a counterexample:
every=.&> NB. uses compose
each=.&.> NB. uses under
rule =: (f every) -: >@(f each)
NB. Is completely general?
thank you, Dave
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