Re: Idempotence and "Replication Insensitivity" are equivalent ?

<pamelafluente_at_libero.it> wrote:
> all this discussion was based on my attempt
>
> to show that the implication
>
> "duplication insensitivite" ==> idempotent
>
> does not hold, and now you are saying you never meant it ?

I can't speak for what Marshall meant, but I can tell you he never said it. That's why, when you posted to sci.math to ask whether that statement is true, I clarified that Marshall had never said that statement, and tried to explain what he really said.

> Under the assumption that exists f2 such that
> for each n>2 we can write fn (assumed associative/commutative) as:
>
>
> f3(x,y,z) = f2( f2(x,y), z)
>
> ...
>
> fn(x1,...xn) = fn-1( fn-1(x,y), xn)
>
> we have:
>
> f2 idempotent ==> fn "duplication insensitivite"

Yes, and the implication goes the other way, as well. However, you have a slight problem in that your function is undefined on fewer than two values.

> under the assumption that fn is "duplication insensitivite"
> and n>3,
>
> we can never imply f2(x,y) idempotent because
> duplication insensitivity can be attained
> duplicating item that are different among them,
> eg. 3,3,5,5
>
> while idempotence requires at each "binary step"
> equal arguments.

Idempotence requires that *if* the arguments are equal, then the result will be equal to the argument. An idempotent function can certainly be evaluated with unequal arguments. Furthermore, proofs have been given in this thread of the very thing that you are saying can never be implied. (The counterexamples come in when you define things a bit differently; the model you are talking about is still the limited one that cannot define COUNT; in that case, there really is a complete equivalence.)

-- 
Chris Smith