(f g h) is a fork, where the last verb executed is g. the result after g is
executed is the same as the result after the fork is executed, as they are both
the same moment.
A1 =: 1 : 'w@:u'
A2 =: 1 : 'w m'
will produce the same noun results in: (where y is noun right argument, f g h
w are verbs) n: is the proposed adverb that makes this equivalence possible.
(f g h)A1 y
(f g A1 h) y
((f g h) y) A2
((f g h) n: A2) y
(f g n: A2 h) y NB. (n: A2) does not change "fork nature".
On Saturday, January 14, 2023 at 03:15:19 a.m. EST, Elijah Stone
<elro...@elronnd.net> wrote:
(By coincidence, I mean that uACv is not the same as uCvA, in general. Nor
uA1A2 the same as uA2A1. And f A g h is not the same as (f g h)A, either,
e.g.)
On Sat, 14 Jan 2023, Elijah Stone wrote:
> I cannot make heads nor tails of anything you have said. That f g(u@:) h is
> the same as (f g h)(u@:) is true, but coincidence, and I don't see what it
> has
> to do with anything.
>
> On Sat, 14 Jan 2023, 'Pascal Jasmin' via Programming wrote:
>
>> Raul expressed by thinking,
>>
>> .> x (f g n:A h) y would be same as (f g h) n: A -> (x ((x f y) g (x h
>> y)) y)A
>>
>> the logic is that g executes 3rd/last in (f g h), and f g(u@:) h) is same
> as (f g h)(u@:)
>>
>> n: (A =: 1 : 'v m')is similar to (v@:) but applies to the result of the
> verb phrase u (applied to x,y) instead of the verb phrase u
>>
>>
>>
>> On Friday, January 13, 2023 at 09:36:28 p.m. EST, Elijah Stone
> <elro...@elronnd.net> wrote:
>>
>>
>>
>>
>>
>> Not quite (at least, not in my conception of it). If it is to be useful in
> a
>> larger verb train, you have to work out where exactly x and y come from.
> For
>> instance, if we have x (f g n:A h) y, should we apply (x f y) ((x f y) g (x
> h
>> y))A (x h y)? Or (x f y) (x g y)A (x h y)? I say it should be chosen in
> the
>> same way as $: (which leads to the latter in this case).
>>
>> On Fri, 13 Jan 2023, Raul Miller wrote:
>>
>>> I find it difficult to reason about this n:
>>>
>>> My best guess is that n: is itself an adverb and that u n: A (where u
>>> is a verb and A is an adverb) would be handled by special code which
>>> behaves like
>>> {{ (u y) A}} : {{(x u y) A}}
>>>
>>> Does that agree with your thinking?
>>>
>>> Thanks,
>>>
>>> --
>>> Raul
>>>
>>> On Fri, Jan 13, 2023 at 7:38 PM 'Pascal Jasmin' via Programming
>>> <programm...@jsoftware.com> wrote:
>>>>
>>>> To answer Raul, I did not use r2m after all. oa through the magic of
> cloak allows 'Adverb' oa ('X' oa in example) where Adverb has a noun
> parameter.
>>>>
>>>> > I had: u n: A y is (u y) A y. Whereas you have u r2m A y as simply (u
> y) A.
>>>>
>>>> if [x] u n: A y produced the result of x u y as input to A, then that is
> a legal way to get Adverb noun inputs from a verb phrase. An adverb can
> create modifiers is the main benefit, and necessity for the functionality.
>>>>
>>>>
>>>> I feel that u n: A y as (u y) A y would be for producing verbs and noun
> results, and can be written as 1 : '(u y) A y' though that doesn't let you
> produce a conjunction from A and return (C y).
>>>>
>>>> If there is ever an attack on the supreme majesty that is Cloak, I do
> hope n: is implemented instead.
>>>>
>>>>
>>>> On Friday, January 13, 2023 at 05:39:30 p.m. EST, Elijah Stone
> <elro...@elronnd.net> wrote:
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> Oh, my n: is a little less expressive than your r2m. I had: u n: A y is
> (u y)
>>>> A y. Whereas you have u r2m A y as simply (u y) A.
>>>>
>>>> On Fri, 13 Jan 2023, Elijah Stone wrote:
>>>>
>>>> > I proposed your 'r2m' as a primitive n: (for 'now') a while ago, and
> received
>>>> > a lukewarm response. I don't think it can be implemented other than as
> a
>>>> > primitive. (And I still think it would be a good idea to have.)
>>>> >
>>>> > Your solution which quotes the modifier name works, but I find it
>>>> > distasteful.
>>>> > And it has some trouble with conjunctions; how do you disambiguate the
>>>> > following?
>>>> >
>>>> > (u r2m) C v
>>>> >
>>>> > u C (v r2m)
>>>> >
>>>> > (u r2m) C (v r2m)
>>>> >
>>>> > You can't, so you would need a separate form for each.
>>>> >
>>>> > On Fri, 13 Jan 2023, 'Pascal Jasmin' via Programming wrote:
>>>> >
>>>> >> X =: 1 : 'm&+'
>>>> >>
>>>> >>
>>>> >> What definition of r2m (result to m argument) below would allow X to
> see
>>>> > the result of + y (or x+y) as its m argument?
>>>> >>
>>>> >> + r2m X 3
>>>> >>
>>>> >> purpose would be for X to produce a modifier from application of
> "verb".
>>>> > Requirement is only that y argument (3 above) is outside any verb
> phrase.
>>>> >>
>>>> >> Jose/Dan's Cloak magic? turn result into atomic or linear
> representation?
>>>> >> ----------------------------------------------------------------------
>>>> >> For information about J forums see http://www.jsoftware.com/forums.htm
>>
>>>>
>>>> > ----------------------------------------------------------------------
>>>> > For information about J forums see http://www.jsoftware.com/forums.htm
>>>> >
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
