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la pycyn cusku di'e
I agree that it ought not be the case, but where exactly do my proofs that itdoes follow fail?
They fail only when you treat {lo broda} as an individual term.
As near as I can make out your derivation trying to be fair to you, it goes
like this:
busku lo'e broda = sisku tu'o ka ce'u broda (original) Def 2
= sisku tu'o ka da poi broda zo'u ce'u du da (a
questionable move, though extensionally OK)
= sisku tu'o ka ce'u du lo broda (ditto)
All correct.
In any case, at no point is it available to you to get to {da poi broda zo'usisku tu'o ka ce'u du da}.
Of course! That's the whole point. You can't get to that from
{buska lo'e broda} because that's a different meaning.
So, if you did not treat {lo} as an individual
term you made one and perhaps another of several questionable to clearly
illegal moves.
No. This is the definition of buska:
\x\y buska(x,y) = \x\y sisku(x, \z du(z,y) )
Now from that definition, you can see that:
da poi broda zo'u buska da
= da poi broda zo'u sisku tu'o ka ce'u du da
This can also be written as:
buska lo broda
= da poi broda zo'u sisku tu'o ka ce'u du da
As you can see, no questionable moves there.
Sorry, it just doesn't work the way you think it should -- or the way I
think
it should as far as {lo'e} and {lo} go, though I don't agree with the other
pieces (especially since they make for trouble).
If it doesn't work, you haven't shown why. mu'o mi'e xorxes _________________________________________________________________Surf the Web without missing calls!�Get MSN Broadband. http://resourcecenter.msn.com/access/plans/freeactivation.asp