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Nick:
#And that is true; only one quarter of the Beatles is called George:
#pa lo fi'u ro loi prenrbitlzi cu se cmene zo djordj
#pa lo prenrbitlzi cu se cmene zo djordj
#
#at least one quarter of the Beatles weighs less than
#100 kg. In fact, all the quarters do (did):
#
#pa lo fi'u ro loi prenrbitlzi cu ki'ogra li su'e 100
#pa lo prenrbitlzi cu se ki'ogra li su'e 100
{su'o pa lo} or {ro lo}
#(3) mi djica lenu mi tavla fi'u vo loi prenrbitlzi
#i.e.
#mi djica lenu mi tavla tu'o lo fi'u ro loi prenrbitlzi
#mi djica lenu mi tavla tu'o lo prenrbitlzi
#
#Under this interpretation, there isn't necessarily anything to go to a
#prenex when you see pimu loi or fi'u ro loi. In (1), what would have to
#go to a prenex as an overt outer quantifier is pa da. In (2), it is ro
#da. In (3), nothing goes to the (outermost, extensional) prenex at all:
#it is tu'o da.
#
#Therefore, the Lojban lojbanmass loi broda (which is always implicitly
#quantified) includes in its denotation Mr broda. In particular, fi'u ro
#loi broda can mean Mr Single Broda, and pisu'o broda means Mr Any
#Number of Broda = Mr Broda (since pisu'o >= fi'u ro).
because, if I understand you, {piPA loi} = {Q lo piPA loi}, where Q can
be tu'o. And tu'o really is a zi'o-type quantification killer.
By the same token, {lo broda} is {Q lo broda} and hence can be
{tu'o lo broda} and also give us the kind, after glorking. And {tu'o
broda}, being short for {tu'o lo broda}, can also give us the kind,
without glorking..
I wonder how we quantify over members of kinds and over subkinds
of kinds. Aha:
members: PA lo fi'u ro loi tu'o lo broda
subkinds: PA lo ci'ipa lo tu'o lo broda [maybe. But maybe not.]
but those are craply unthought-out. Plus, I don't see why kinds can't
have bits, so we also need to be able to quantify over bits of
kinds (though perhaps it is redundant, since anything that is a
bit of Mr Broda is presumably also a bit of broda).
--And.