u&.v y has the rank of the monad v.
The Dictionary is misleading about this. Actually you have
x u&.:v y ↔ vi (v x) u (v y)
and then
u&.v ↔ u&.:v"({. v b. 0)
Henry Rich
On 10/26/2011 7:18 AM, Linda Alvord wrote:
> Sorry, Wrong rule.
>
> u&.v y ↔ vi u v y
>
>
> From: [email protected] [mailto:programming-
> [email protected]] On Behalf Of Linda Alvord
> Sent: Wednesday, October 26, 2011 7:12 AM
> To: 'Programming forum'
> Subject: [Jprogramming] problem with under
>
> x u&.v y ↔ vi (v x) u (v y)
>
>
>
> ([: /: ":)&.>a,b
>
> ┌─────────┬─────────┐
> │1 0 2 3 4│4 2 3 1 0│
> └─────────┴─────────┘
>
> <([: /: ":)>a,b
>
> ┌──────────────────────┐
> │5 1 10 0 2 8 9 3 7 4 6│
> └──────────────────────┘
>
>
>
> Why are these different? Linda
>
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