Re: Idempotence and "Replication Insensitivity" are equivalent ?

pamelafluente_at_libero.it wrote:
> Chris Smith ha scritto:
>
> > Actually, f(0,0)=0 and f(1,1)=1, so actually f is idempotent. The value
> > of f(0,1) is not relevant to idempotence.
>
>
> Simply replacing AND with NOR seems to work
>
> Let's try this patch to the previous counterexample.
>
> ----------------------------------------------------------------
>
> Replace the product f with the NOR function:
>
> NOR
> yields false if any condition is true
> http://mathworld.wolfram.com/NOR.html
>
> f(T,T) = F
> f(T,F) = F
> f(F,T) = F
>
> f(F,F) = T
>
>
> 1.
> we have f(F,F) = T
>
> So f is NOT idempotent

Correct.

> 2.
> f is "duplication insensitive".

Incorrect.

Consider {|T, F|}

Nor folded over {|T, F|}

• NOR(T, F)
• F

Now duplicate the T element: {|T, T, F|}

fold(NOR, {|T, T, F|})

• NOR(NOR(T, T), F)
• NOR(F, F)
• T

Marshall Received on Tue Sep 19 2006 - 20:37:36 CEST