Re: Idempotence and "Replication Insensitivity" are equivalent ?

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

 2.
   f is "duplication insensitive". Infact:

   f can take values only in {F, T}

   2A.

   if f(.) = F,

   this result cannot be changed by any further operands    because to have f(.) = F it is sufficient to have 1 operand equal    to T, and, once we have it, the result cannot turn to T anymore

   2B.

   if f(.) = T,

   it means that all operands are F (because just 1 T would turn the result to F).

   Hence the duplication of an operand (a new F) does not change the result.

   "NOT idempotent" and "duplication insensitivity" co-exist.

   Hence "duplication insensitivity" => idempotence does not hold.

    Got it right ?

-P Received on Tue Sep 19 2006 - 19:48:53 CEST