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I prefer Jorge's position. I know that's not a very helpful contribution
from me, but sometimes it helps to know where people stand on
issues.
--And.
>>> Jorge Llambias <jjllambias@hidden.email> 03/06/02 12:27am >>>
la pycyn cusku di'e
>Lojban is now on the brink of being able to use the complete set of these
>quantifiers: the + group is {Q (lo) broda cu brode}, the - group is {Q da
>poi
>broda cu brode}.
There is another way to do it:
A+ ro lo su'o broda cu brode
E+ no lo su'o broda cu brode
I+ su'o lo su'o broda cu brode
O+ me'iro lo su'o broda cu brode = da'asu'o lo su'o broda cu brode
A- ro lo [ro] broda cu brode
E- no lo [ro] broda cu brode
I- su'o lo [ro] broda cu brode
O- me'iro lo [ro] broda cu brode
I can't really believe that {su'o da poi broda} is I-, true
in the absence of broda, but if that works, so should {su'o
lo ro broda}. Same for O-.
More credible O- and I- are:
O- naku ro lo su'o broda cu brode
I- naku no lo su'o broda cu brode
On the other hand, A+ and E+ are not at all controversial
as {ro lo su'o broda} and {no lo su'o broda}.
>Assuming that {ro} and {su'o} behave properly for A+, A-
>and I+ and that {no} works for E+ and E- and that O+ is just {su'o S cu
>naku
>P}, we need only a new form for O-. {na'e ro} fills the bill, for even if
>S
>is empty, the value will be different from {ro}.
But {na'e ro} is not a grammatical quantifier.
mu'o mi'e xorxes
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