I think what you are looking for is the definition of the &
conjunction. You need to understand what it means to be a verb defined
that way.

And, for this, you need to read the reference documentation:

http://www.jsoftware.com/help/dictionary/d630n.htm states:

      x m&v y <---> m&v^:x y

Here, x, m and y are nouns and v is the base verb. Let's call this "Rule A"

It also states:

      x u&n y <---> u&n^:x y

Here, x, n and y are nouns and u is the base verb.  Let's call this "Rule B"

(I say "base" verb, because u&n is a composite verb, derived from the
dyadic definition of u with n as its right argument.)

-------------------

But it also states:

      m&v y is defined as m v y

and

      u&n y is defined as y u n

Let's call these "Rule C" and "Rule D".

-------------------

In other words,

   ((2x&*) &1) 3

Rule D works with u: (2x&*), n: 1, y: 3, and gives us

   3 (2x&*) 1

Rule B works with x: 3, u: (2x&*), y: 1

   ((2x&*)^:3) 1

By the way, I should probably note that we need the outer parenthesis
there to prevent the 3 and 1 from being interpreted as a single token.
But there are a couple other ways we could have done that:

   (2x&*)^:(3) 1

or

   (2x&*)^:3 (1)

We could also have used an expression to generate the value 1. For example:

   (2x&*)^:3 ]1

or

   (2x&*)^:3 >:0

-------------------

Does that make sense?

Thanks,

-- 
Raul

On Thu, Feb 19, 2015 at 9:08 AM, Fausto Saporito
<[email protected]> wrote:
> Hello Raul,
>
> thanks for the explanation. I'm still trying to understand it, but I
> thini I got it.
>
> Sorry my mistake about missing parenthesis.
>
> So, I'm missing the rule to understand this passage:
>
>  ((2x&*) &1) 3  <---> ((2x&*)^:3) 1
>
> But maybe this subject is too much advance for my J actual knowledge.
>
> thanks,
> Fausto
>
>
> 2015-02-19 14:48 GMT+01:00 Raul Miller <[email protected]>:
>> Please be careful here:
>>
>>    3 ((2x&*) &1) 3
>>
>> is equivalent to each of these:
>>    ((2x&*) &1) ((2x&*) &1) ((2x&*) &1) 3
>>    (((3(2x&*)1) (2x&*)1) (2x&*)1)
>>    ((((2x&*)(2x&*)(2x&*)1) (2x&*)1) (2x&*)1)
>>    ((8 (2x&*)1) (2x&*)1)
>>    (((2x&*)(2x&*)(2x&*)(2x&*)(2x&*)(2x&*)(2x&*)(2x&*)1) (2x&*)1)
>>    (256 (2x&*)1)
>>
>> and I'm not going to carry out the final step, which involves 256
>> copies of the verb 2x&*
>>
>> Hopefully you can see why each of these expressions is equivalent. But
>> if something doesn't make sense, it's probably worth talking through
>> the issue (or at least showing a more gradual sequence of equivalences
>> for that step).
>>
>> On the other hand,
>>    ((2x&*) &1^:3 3
>>
>> is not a valid expression because of the unbalanced parenthesis.
>>
>> If you got rid of one of those left parenthesis, you'd get
>>    (2x&*) &1^:3 3
>> which is equivalent to
>>    (2x&*) &1^:(3 3)
>> but that is a verb which you probably did not intend.
>>
>> If you instead insert a right parenthesis between the pair of threes,
>> you would get
>>    ((2x&*) &1^:3) 3
>>
>> which indeed is equivalent to your first expression.
>>
>> But note that you are not repeating three times 2x&* but instead are
>> repeating three times ((2x&*)1) in much the same manner as the
>> original expression.
>>
>> Thanks,
>>
>> --
>> Raul
>>
>>
>> On Thu, Feb 19, 2015 at 8:11 AM, Fausto Saporito
>> <[email protected]> wrote:
>>> Hello Jose,
>>>
>>> I understand you are applying this identity : x u&n y <--> u&n^:x y
>>>
>>> But I cannot see the correct mapping in your expressions.
>>> I suppose this is an hook, so
>>>
>>> (u v) y  -->  y u v y
>>>
>>> 3 ((2x&*) &1) 3
>>>
>>> ((2x&*) &1^:3 3
>>>
>>> means repeat three times 2*1, i.e. 2*1*2*1*2*1 = 8
>>>
>>> correct ?
>>>
>>> thanks,
>>> Fausto
>>>
>>>
>>> 2015-02-19 1:19 GMT+01:00 Henry Rich <[email protected]>:
>>>> If you're going to use dissect, get 3.6.42 (released today).  Previous
>>>> versions had a confusing title for the verbs.
>>>>
>>>> Even with the picture it's amazing what this little phrase does.  Two 
>>>> nested
>>>> powers, with the result of one power feeding back into the exponent of the
>>>> next iteration of the same verb.
>>>>
>>>> Henry Rich
>>>>
>>>>
>>>> On 2/18/2015 7:12 PM, Jose Mario Quintana wrote:
>>>>>
>>>>>     ((2x&*) &1) 3
>>>>> 8
>>>>>     ((2x&*)^:3) 1
>>>>> 8
>>>>>     ((3x&*) &1) 2
>>>>> 9
>>>>>     ((3x&*)^:2) 1
>>>>> 9
>>>>>
>>>>>
>>>>> Does dissect
>>>>>
>>>>>
>>>>> http://www.jsoftware.com/jwiki/action/show/Vocabulary/Dissect?action=show&redirect=Addons%2Fdebug%2Fdissect
>>>>>
>>>>> help to follow the execution of the sentences?
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
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>> ----------------------------------------------------------------------
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