Re: ?? Functional Dependency Question ??

On Oct 27, 9:36 am, JOG <j..._at_cs.nott.ac.uk> wrote:
> Apologies for double click posting.
>
> On Oct 23, 2:23 am, David BL <davi..._at_iinet.net.au> wrote:
>
> > On Oct 22, 7:20 pm, JOG <j..._at_cs.nott.ac.uk> wrote:
> > > Ax [ inRoom(x) ^ isHippo(x) ^ Wearing(x, pink panties) ] = false
>
> > Is this actually the case? There doesn't appear to be any defined set
> > over which the universal quantification is defined, so I think the
> > left hand side is meaningless not false.
>
> While you're right, could you not read that first statement as:
> Ax [Exists(x) -> inRoom(x) ^ isHippo(x) ^ Wearing(x, pink panties) ]

Assuming that someone has properly defined these relations, and in particular Exists(x) (or equivalently its extension is well defined), that translation would presumably be false.

> How do you like them apples! Either way, false or meaningless, its
> clearly not what the lecturer was saying so the second statement was
> the banker:
> Ax [inRoom(x) ^ isHippo(x) -> Wearing(x, pink panties) ]
>
> which is true (material implication is always true if the antecedent
> is false). Hence, as far as formal logic is concerned all the hippos
> in the room are indeed wearing pink panties. And blue panties too in
> fact. And no panties as well...
>
>
>
>
>
> > I think a meaningful universal quantification must be able to be
> > written in the form
>
> > Ax [ P(x) -> Q(x) ]
>
> > where { x | P(x) } is a well defined set.
>
> > We can write
>
> > Ax [ inRoom(x) ^ isHippo(x) ^ Wearing(x, pink panties) ]
>
> > as
>
> > Ax [ true -> inRoom(x) ^ isHippo(x) ^ Wearing(x, pink panties) ]
>
> > So it seems that we must define P(x) = true, but then { x | P(x) } is
> > the set of all things which is meaningless.
>
> Why is the "set of all things" meaningless?

An axiom of set theory states that it is always meaningful to define a subset of a given set according to some well defined predicate.

ie the axiom states that given set S, and predicate f, { x in S | f(x) } is a well defined set.

In particular we can take the subset of the extension of P(x) defined as the set of all sets that are not members of themselves, leading to Russell's paradox. Received on Tue Oct 28 2008 - 05:27:06 CET