As has been written here several times, when the Dictionary says

u&.v y ↔ vi u v y

it is wrong.  [Others hold that it is right but that the statement needs
to be viewed in light of the ranks shown in the Dictionary.]

Either way, you just can't make that substitution and get right results.
 You can either use the forms that I have given, or you can understand
rank as Kip, Dan, Raul, & others have proposed.

If you are appealing to the false equivalence shown above (or the
similar ones for @ and &), when you get wrong results it's because those
equivalences do not take rank into account.

Henry Rich

On 1/22/2013 2:07 AM, Linda Alvord wrote:
> Here’s another puzzle for me:
> 
>      NB. u&.v y ↔ vi u v y
>     i:&.> +. 1j2
> ┌──────┬───────────┐
> │_1 0 1│_2 _1 0 1 2│
> └──────┴───────────┘
>   
>     ]A=:+.1j2
> 1 2
>     <i:> A
> ┌───────────┐
> │_1  0 1 0 0│
> │_2 _1 0 1 2│
> └───────────┘
> 
> I must be using the definition incorrectly.
> 
> Linda
> 
> -----Original Message-----
> From: [email protected] [mailto:programming-
> [email protected]] On Behalf Of David Ward Lambert
> Sent: Monday, January 21, 2013 10:45 AM
> To: programming
> Subject: Re: [Jprogramming] [Jprogrammingou Hermitian from triangular
> 
> Interrupting may be a bad idea.
> 
>     NB. -------- Use 13 : correctly.
>     NB.    f=: 13 :',j./&i:/+.y'           NB. no.
>     NB.    f=: 13 : '([: , j./&i:/@+.) y'  NB. yes.
> 
>     ic2 =: [: , j./&i:/@+.
>     
>     NB. Using a proverb is as if its definition
>     NB. were inserted with parenthesis.
>     NB. Copy ic2 definition verbatim,
>     NB. Surround with parentheses.
>     NB. Put in the arguments, then fix quotes.
> 
>     13 : '([: , j./&i:/@+.) y'
> [: , j./&i:/@+.
> 
>     NB. The definition was tacit,
>     NB. 13 : returns this tacit verb unchanged
> 
> 
> 
>     NB. -------- Equivalence of ic and ic2
>     NB."However, I do not understand how ic and ic2 agree when they don't!"
>     
>     ic =: [: , ([: i: 9&o.) j./ [: i: 11&o.
> 
>     ic2 =: [: , j./&i:/@+.
> 
>     NB. ic  and  ic2  are identical at rank 0
>     (ic -: ic2)&>1j2 0j2
> 1 1
> 
>     ([: < ic)"0 ] 1j2 0j2
> +----------------------------------------------------------------+----------
> ---------+
> |_1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2|0j_2
> 0j_1 0 0j1 0j2|
> +----------------------------------------------------------------+----------
> ---------+
>     ([: < ic2)"0 ] 1j2 0j2
> +----------------------------------------------------------------+----------
> ---------+
> |_1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2|0j_2
> 0j_1 0 0j1 0j2|
> +----------------------------------------------------------------+----------
> ---------+
> 
> 
> 
>> Date: Mon, 21 Jan 2013 03:58:24 -0500
>> From: "Linda Alvord" <[email protected]>
>> To: <[email protected]>
>> Subject: Re: [Jprogramming] [Jprogrammingou Hermitian from triangular
>> Message-ID: <001d01cdf7b5$78a4cd20$69ee6760$@net>
>> Content-Type: text/plain;       charset="koi8-r"
>>
>> Since this has a mistake it is hard to reconstruct what I did wrong.
> This
>> represents what I was doing:
>>
>>
>>
>>    ic2=: [: , j./&i:/@+.
>>
>>    ic2
>>
>> [: , j./&i:/@+.
>>
>>    ([: , j./&i:/@+.)2
>>
>> _2 _1 0 1 2
>>
>>    f=: 13 :',j./&i:/+.y'
>> ................
>>
>>    ic 1j2 0j2
>>
>> _1j_2 _1j_1 _1 _1j1 _1j2 _1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2
> 0j_2
>> 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2 1j_2 1j_1 1 1j1 1j2 0j_2 0j_1 0 0j1
> 0j2
>> 0j_2 0j_1 0 0j1 0j2 0j_2 0j_1 0 0j1 0j2 0j_2 0j_1 0 0j1 0j2 0j_2 0j_1 0
> 0j1
>> 0j2 0j_2 0j_1 0 0j1 0j2
>>
>>    
>>
>>    ic2 1j2 0j2
>>
>> _1j_2 _1j_1 _1 _1j1 _1j2 0j_2 0j_1 0 0j1 0j2 1j_2 1j_1 1 1j1 1j2 0j_2
> 0j_1 0
>> 0j1 0j2 0 0 0 0 0 0 0 0 0 0
>>
>>   
>>
>> (ic-:ic2)&>1j2 0j2
>>
>> 1 1
>>
>>    
>>
>> However, I do not understand how ic and ic2 agree when they don't!
> 
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