Re: [jboske] Llamban (was Re: Lo'e le'e

["Jorge Llambias" <jjllambias@hotmail.com>]

la pycyn cusku di'e

> I agree that it ought not be the case, but where exactly do my proofs that 
> it
> does 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'u
> sisku 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



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