It already tells us that, just not on that page.

If you visit the Vocabulary page, and then hit the << link at the top,
you get this page:

http://www.jsoftware.com/help/dictionary/dict3.htm

Note that all of section I (Alphabet) and II (Grammar) are also relevant.

-- 
Raul

On Sat, Jan 12, 2013 at 9:51 AM, Linda Alvord <lindaalv...@verizon.net> wrote:
> Here's a sentence that I think would be more helpful in the vocabulary:
>
>      ]A=:'one';'two';'three'
> ┌───┬───┬─────┐
> │one│two│three│
> └───┴───┴─────┘
>    #&>A
> 3 3 5
>    ([:#>)">A
> 3 3 5
>                     ^
>    NB. Then shouldn't the vocabulary say   u&v y ->  ([: u v y)"v
>
> A similar sentence for  @  might save beginners a lot of anguish.
>
> Linda
>
> -----Original Message-----
> From: programming-boun...@forums.jsoftware.com 
> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of km
> Sent: Saturday, January 12, 2013 4:20 AM
> To: programm...@jsoftware.com
> Subject: Re: [Jprogramming] joining to an empty list
>
> You have quoted the vocabulary entry correctly, but must interpret the first 
> sentence in the light of the third.  (Look at the entry again and count 
> periods to locate the first and third sentences.)
>
> Your usual tools can help:
>
>     f =: 13 : '#&> y'
>     f
>  #&>
>     g =: 13 : '# > y'
>     g
>  [: # >
>
> Because > has ranks 0 0 0 and monad #&> is equivalent to #@>, monad #&> is 
> equivalent to
>
>     h
>  ([: # >)"0
>
>     f 'one';'two';'three'
>  3 3 5
>     h 'one';'two';'three'
>  3 3 5
>     g 'one';'two';'three'
>  3
>
> Kip Murray
>
> Sent from my iPad
>
>
> On Jan 12, 2013, at 1:37 AM, "Linda Alvord" <lindaalv...@verizon.net> wrote:
>
>> My issue is with the vocabulary. Here is compose:
>>
>> u&v y ↔ u v y . Thus +:&- 7 is _14 (double the negation). Moreover, the 
>> monads u&v and u@v are equivalent.
>>
>>   f=: 13 :'#&>y'
>>    f 'one';'two';'three'
>> 3 3 5
>>
>>   g=: 13  :'#>y'
>>   g 'one';'two';'three'
>> 3
>>
>> I play the part of a perpetual beginner.  When a statement is in the 
>> vocabulary I should be able to trust that it will work.  As I see it, it 
>> does not work.
>>
>> My question is why not?
>>
>> Linda
>>
>> -----Original Message-----
>> From: programming-boun...@forums.jsoftware.com 
>> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Linda Alvord
>> Sent: Saturday, January 12, 2013 2:19 AM
>> To: programm...@jsoftware.com
>> Subject: Re: [Jprogramming] joining to an empty list
>>
>> Fortunately it shows up in my sent messages in small print.
>>
>> Linda
>>
>> -----Original Message-----
>> From: programming-boun...@forums.jsoftware.com 
>> [mailto:programming-boun...@forums.jsoftware.com] On Behalf Of Raul Miller
>> Sent: Friday, January 11, 2013 10:14 PM
>> To: programm...@jsoftware.com
>> Subject: Re: [Jprogramming] joining to an empty list
>>
>> On Fri, Jan 11, 2013 at 10:07 PM, Linda Alvord <lindaalv...@verizon.net> 
>> wrote:
>>> I don't understand why  f  and  g  do not agree?
>>
>> Because they are doing something different.
>>
>>>   #&>'one';'two';'three'
>>> 3 3 5
>>>   f=: 13 :'#&>y'
>>>   f 'one';'two';'three'
>>> 3 3 5
>>
>> f counts the number of items in each box.
>>
>>>   NB. Compose:   u&v y ↔ u v y
>>>
>>>  g=: 13  :'#>y'
>>>   g=: 13  :'#>y'
>>>   g 'one';'two';'three'
>>> 3
>>
>> g counts the number of items after unboxing.
>>
>> Let's try replacing # with <
>>
>>   <@> 'one';'two';'three'
>> +---+---+-----+
>> |one|two|three|
>> +---+---+-----+
>>   <> 'one';'two';'three'
>> +-----+
>> |one  |
>> |two  |
>> |three|
>> +-----+
>>
>> Or, let's change the data and use $
>>
>>    $@> i.each 2 2; 3 3; 4 4
>> 2 2
>> 3 3
>> 4 4
>>   $> i. each 2 2; 3 3; 4 4
>> 3 4 4
>>
>> Remember that > has rank 0 so F@> is going to run an independent instance of 
>> F for each box that you are unpacking.
>>
>> Does this help?
>>
>> --
>> Raul
>> ----------------------------------------------------------------------
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>>
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