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Home -> Community -> Usenet -> comp.databases.theory -> Re: circular relationships ok?
"JOG" <jog_at_cs.nott.ac.uk> wrote in message
news:1141494992.645738.283070_at_z34g2000cwc.googlegroups.com...
> David Cressey wrote:
> > "JOG" <jog_at_cs.nott.ac.uk> wrote in message
> > news:1141480471.362482.59010_at_e56g2000cwe.googlegroups.com...
> > > David Cressey wrote:
> > > [snip]
> > > >
> > > > I think "A implies B" is the same as "B or not A".
> > > >
> > >
> > > ? By "A implies B", he surely just means "if A then B". Or using
> > > standard predicate logic notation "A -> B."
> >
> > I don't get it. How is "A -> B" different from "B^~A" ?
> >
>
You are right. It was a typo. Thanks for correcting it.
> A -> B is a proposition
> B v ~A is a boolean expression
>
It may be obvious to you, but it ain't obvious to me. It seems to me that
the assertion
that A -> B is always true in a given univers of discourse, and the
assertion that
B is true or A is false in the same universe of discourse boil down to the
same thing.
> So I thought that you might perhaps have meant "B v ~A = True", but
> substituting a couple of values in for A and B highlights there still
> exists a difference:
>
>
>
I don't think the above analysis is correct. Your analysis requires OR to
mean "exclusive or".
It doesn't. It means inclusive or, doesn't it?
In the case where B = I will wear a coat and A= it is raining Then if I wear a coat and it's not raining we get:
B v ~A becomes True v not false becomes True v true becomes true. Right?
> All best, Jim.
>
Received on Sat Mar 04 2006 - 12:18:58 CST
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