That looks right, to me.

Note that you can test this if you replace := with -:

For example:

x=: 1
y=: 2
a=:B=:c=:D=:e=:F=: <@,&<

Thanks,

-- 
Raul


On Sun, Jul 21, 2013 at 7:53 PM, Tobia Conforto
<[email protected]> wrote:
> Hello
>
> I'm trying to understand trains of verbs and I came up with this. Can anybody 
> please either confirm it or correct me?
>
> Monadic trains:
>
>             (B a) y  :=                 y  B (a y)
>           (c B a) y  :=              (c y) B (a y)
>         (D c B a) y  :=         y  D (c y) B (a y)
>       (e D c B a) y  :=      (e y) D (c y) B (a y)
>     (F e D c B a) y  :=  y F (e y) D (c y) B (a y)
>
> Dyadic trains:
>
>           x (B a) y  :=                     x  B (  a y)
>         x (c B a) y  :=                (x c y) B (x a y)
>       x (D c B a) y  :=           x  D (  c y) B (  a y)
>     x (e D c B a) y  :=      (x e y) D (x c y) B (x a y)
>   x (F e D c B a) y  :=  x F (  e y) D (  c y) B (  a y)
>
> Capped fork in even-numbered dyadic train:
>
> x (F e  D c  B a) y  :=  x F (e y) D (c y) B (a y)
> x (F e  D [: B a) y  :=  x F (e y) D       B (a y)
> x (F [: D c  B a) y  :=  x F       D (c y) B (a y)
> x (F [: D [: B a) y  :=  x F       D       B (a y)
>
> Capped fork in odd-numbered dyadic train:
>
>   x (e  D c  B a) y  :=  (x e y) D (x c y) B (x a y)
>   x (e  D [: B a) y  :=  (x e y) D         B (x a y)
>   x ([: D c  B a) y  :=          D (x c y) B (x a y)
>   x ([: D [: B a) y  :=          D         B (x a y)
>
>
> -Tobia
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