Re: [lojban] xoi and new soi as bridi relative clause

[Ilmen <ilmen.pokebip@gmail.com>]

On 26/07/2015 15:29, guskant wrote:

Le jeudi 23 juillet 2015 21:56:28 UTC, Ilmen a écrit :

Please correctly if I made any mistake:

ro da poi ke'a su'o de viska cu se kanla su'o di
= ro da poi su'o de zo'u ke'a de viska ku'o su'o di zo'u di da kanla
= ro da zo'u ganai su'o de zo'u da de viska gi su'o di zo'u di da kanla
= roldza fa loka ko'a ce'ai ganai suzdza fa loka ko'a ce'u viska gi suzdza fa loka ce'u ko'a kanla
= roldza fa loka ko'a ce'ai zilvlina fa lodu'u jitfa fa lodu'u suzdza fa loka ko'a ce'u viska kei kei kei fe lodu'u suzdza fa loka ce'u ko'a kanla

With this expansion, you can see that actually da/de/di outside of a prenex expand to {ce'u}, so it is an open slot used for making properties of which we will quantify how many entities they are satisfied with, using logical quantification predicates (roldza, suzdza).

mi'e la .ilmen. mu'o

It seems correct. And then how do you transform it into a prenex normal form?

Once again, please correct me if I made any mistake:

{ro da zo'u ganai su'o de zo'u da de viska gi su'o di zo'u di da kanla}
= ∀x((∃y(viska(x,y))) → (∃z(kanla(z,x)))

# Application of the "(∃xϕ) → ψ" ≡ "∀x(ϕ → ψ)" transformation:
≡ ∀x∀y(viska(x,y) → (∃z(kanla(z,x)))

# Application of the "ϕ → (∃xψ)" ≡ "∃x(ϕ → ψ)" transformation:
≡ ∀x∀y(∃z(viska(x,y) → kanla(z,x))

= {ro da ro de su'o di zo'u ganai da de viska gi di da kanla} (prenex normal form)

mi'e la .ilmen. mu'o

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