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A pair of how-do-i-say-it's
- To: John Cowan <cowan@SNARK.THYRSUS.COM>,       Eric Raymond <eric@SNARK.THYRSUS.COM>,       Eric Tiedemann <est@SNARK.THYRSUS.COM>
 
- Subject: A pair of how-do-i-say-it's
 
- From: Ivan A Derzhanski <cbmvax!uunet!COGSCI.ED.AC.UK!iad>
 
- Date: Tue, 24 Mar 1992 22:37:28 GMT
 
- In-Reply-To: Chris Handley's message of Wed,             25 Mar 1992 10:05:20 GMT+1200 <5715.9203242203@cogsci.ed.ac.uk>
 
- Reply-To: Ivan A Derzhanski <cbmvax!uunet!COGSCI.ED.AC.UK!iad>
 
- Sender: Lojban list <cbmvax!uunet!CUVMA.BITNET!pucc.Princeton.EDU!LOJBAN>
 
>  Date:     Wed, 25 Mar 1992 10:05:20 GMT+1200
>  From: Chris Handley <CHandley@NZ.AC.OTAGO.GANDALF>
>
>  >Facts:
>  >
>  >  president( GB, US) = true   prime_minister( GB, US) = false
>  >  president( JM, UK) = false  prime_minister( JM, UK) = true
>
>  Correct, but "leader"(GB, US) = true (in some sense) and
>               "leader"(JM, UK) = true (in a somewhat different sense)
Why "but"?  I was objecting against the universal quantification.
`forall bu'a [GB bu'a US -> JM bu'a UK]':
   incorrect (you can think of a kajillion of counterexamples).
`exists bu'a [GB bu'a US /\ JM bu'a UK]':
   correct but uninformative (you can think of a kajillion of
   trivial and irrelevant examples).
What you want is something like `there is a bu'a, such that GB bu'a US
and JM bu'a UK, but it is not a trivial one... no... guess again...'
and the answer to the riddle is `lambda (x,y) [x is leader of y]'.
Afraid you can't put this into logic, though.
Ivan