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Date: Tue, 14 Aug 2012 16:48:27 +0100
From: And Rosta <and.rosta@gmail.com>
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Subject: Re: [lojban] &Lang
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.arpis., On 13/07/2012 19:05:
> On Fri, Jul 13, 2012 at 10:31 AM, And Rosta <and.rosta@gmail.com <mailto:=
and.rosta@gmail.com>> wrote:
>     An example (without wanting to be distracted by irrelevant details):
>
>     "Da djuno lo du'u de prami da gi'e de nelci da"
>     djuno(da, &(prami(de, da), nelci(de, da)))
>
>     Task 1: notate this (linearly) so that "da" and "de" each appear only=
 once.
>
>     And then:
>
>     Task 2: Improve the system so that speakers don't have to hold in mem=
ory the arbitrary correspondence between variable & phonological form.
>
>     Will De Bruijn indices deal with Task 1? If they do, I'd be thrilled =
to have learnt summat new.
>
>
> Ah... No. De Bruijn indices do nothing about factoring out common use sit=
es. You might want to look into the SKI (or SK) calculus; it is capable of =
expressing everything that the lambda calculus is.
>
> What you really want is to define a predicate out of djuno, du'u, prami, =
and nelci that has the right meaning and never use "da" or "de" at all, bec=
ause if they appear only once and in obvious positions, they are superfluou=
s.
>
> Here's how I would go about deriving such a predicate, in case it helps:
>
> Recognize that \ko'a. \ko'e. gi'e (prami ko'a ko'e) (nelci ko'a ko'e) is =
a useful abstraction
> This abstraction can be generalized to
> \conn. \b1. \b2. \ko'a. \ko'e. conn (b1 ko'a ko'e) (b2 ko'a ko'e) which r=
educes to SK as follows
> \conn. \b1. \b2. \ko'a. \ko'e. (conn (b1 ko'a ko'e))                     =
            (b2 ko'a ko'e)
> \conn. \b1. \b2. \ko'a. S (\ko'e. conn (b1 ko'a ko'e))                   =
(\ko'e. (b2 ko'a) ko'e)
> \conn. \b1. \b2. \ko'a. S (S (\ko'e. conn) (\ko'e. (b1 ko'a) ko'e))      =
      (b2 ko'a)
> \conn. \b1. \b2. \ko'a. S (S (K conn) (b1 ko'a))                         =
            (b2 ko'a)
>
> \conn. \b1. \b2. \ko'a. (S (S (K conn) (b1 ko'a)))                       =
       (b2 ko'a)
> \conn. \b1. \b2. S (\ko'a. S (S (K conn) (b1 ko'a)))                 (\ko=
'a. b2 ko'a)
> \conn. \b1. \b2. S (S (\ko'a. S) (\ko'a. S (K conn) (b1 ko'a)))          =
  b2
> \conn. \b1. \b2. S (S S (S (\ko'a. S (K conn)) (\ko'a. b1 ko'a)))        =
 b2
> \conn. \b1. \b2. S (S S (S (S (\ko'a. S) (\ko'a. K conn)) b1))           =
  b2
> \conn. \b1. \b2. S (S S S S S (K conn) b1) b2
>
> And now I'm bored, but I think that this was illustrative.

Either I've misunderstood your response, or you've misunderstood the challe=
nge. The challenge is that the same bound variable need never be repeated. =
Sure for any given single sentence containing n different variables you can=
 define an equivalent single predicate with n arguments, but you can't do t=
his for each of the infinitely many sentences in a language.

> Other things you may want to look at: combinatory logic,
> concatenative programming (or stack-based programming), point-free
> style (or tacit programming), and the J programming language.

Thanks for taking the trouble to make the suggestions, but I don't have the=
 requisite intellectual skills to scan outlines of those areas in the hope =
of descrying a notation that doesn't require variable repetition. If you th=
ink one of them provides a notation that doesn't require variable repetitio=
n, it would be very precious to meif you were to explain it to me.

--and.

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