Re: Relational symmetric difference is well defined
From: V.J. Kumar <vjkmail_at_gmail.com>
Date: Wed, 20 Jun 2007 13:57:41 +0200 (CEST)
Message-ID: <Xns995550FF238A3vdghher_at_194.177.96.26>
> indeed equivalent with "(forall y : A(x,y) <-> B(y,z)) and (exists
> y1 : A(x, y1)) and (exists y2 : B(y2, z))".
>
Date: Wed, 20 Jun 2007 13:57:41 +0200 (CEST)
Message-ID: <Xns995550FF238A3vdghher_at_194.177.96.26>
Jan Hidders <hidders_at_gmail.com> wrote in news:1182338460.046186.165360_at_g4g2000hsf.googlegroups.com:
> I think it is quite clear what I said: you take the formula "(forall
> y : A(x, y) <-> B(y, z))" and you add "and (exists y : A(x,y)) and
> (exists y : B(y, z))". That of course gives you "(forall y : A(x, y)
> <-
>> B(y, z)) and (exists y : A(x, y)) and (exists y : B(y, z))" which is
> indeed equivalent with "(forall y : A(x,y) <-> B(y,z)) and (exists
> y1 : A(x, y1)) and (exists y2 : B(y2, z))".
>
Sounds good! It should work now.
> -- Jan Hidders
>
>
Received on Wed Jun 20 2007 - 13:57:41 CEST