Fred,

A note about novices and Dictionary: they do not
mix together well. Dictionary is the ultimate source,
but not necessarily a first, by design.

I believe that your expectation is based on an assumption
that rank is an intrinsic property of a verb and that the
rank conjunction changes that property.
That's a common pitfall, mentioned in "JfC".

The first part is true, but the second is not.
Rank conjunction does not "attach a new rank" or
"modify the rank property", it's just an ordinary
conjunction, the same as you would say

   u=. 3 : 'y+1'&.>
   u i.3
+-+-+-+
|1|2|3|
+-+-+-+

So it's not about ranks and explicit definitions, 
it's about conjuntions and verbs and the rules using 
explicit script (the : 0), which parses up to ending ")".

To discover how would that work even if you don't
have a dictionary, which is better, you can use the
interactive mode and facilities f. (fix) dereferencing names
and b. for rank.

   v=: 3 : 'y+1'

   v"1 f.
3 : 'y+1'"1    NB. so that's how to write a one-liner with rank

   3 : 'y+1' b. 0  NB. rank of explicit verbs is infinite
_ _ _

   w=: 3 : 0
a=. 1
b=. y
a+b
)
   w"1 f.      NB. so that's how to write explicit with rank
3 : 0"1
a=. 1
b=. y
a+b
)


--- Fred Bone <[EMAIL PROTECTED]> wrote:

> > Fred Bone wrote:
> > > Yes; perhaps I wasn't clear. I was talking not about
> > > applying the verb with a given rank, but about defining
> > > it. I eventually remembered the syntax, for example:
> > >   wossname =. 4 : 0 " 1 0
> > > I still can't find where this is documented.

> And I still don't see how you can tell from just the parsing rules that 
> the rank will become the default rank for the verb whose definition is in 
> the following lines. Let alone that this is the way to go about it.

> how is the novice to know that the construct with the : 0 form is not
>   foo =: 3 : 0
>   blah blah blah
>   ) "1 0

> OTOH, if there had been an example in the Dictionary when I was looking 
> for one, I wouldn't have needed to try it ...



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