Re: "Armstrong's axioms" augmentation - help plz

Dan wrote:
>
> Hello Jan,

Hi Dan,

> What is being disputed here? I thought the question from the OP was whether
> the augmentation rule could be expressed as either:
> 1) {X->Y} |= XZ->YZ
> 2) {X->Y} |= XZ->Y
>
> By augmentation of the first form of the rule, we get {XZ->Y}|= {XZ->YZ}.
> That is the extent of the OP's question, as far as I can tell.

Really? Are you saying that OPs "can also be stated as" should be interpreted as "allows me to derive that"? That doesn't make much sense to me. I'm sorry Dan, I'm having a really hard time trying to follow you here.

> I interpreted your answer of *it cannot* to mean that the second inference
> rule expression was invalid (re: "I don't understand why the second rule can
> also be stated by..."). But now that I read closer, I think you are
> asserting something different.

Indeed I am. Armstrong's rules are an inference mechanism. Given a set of FDs they allow you to derive more FDs, and, as you know, the nice thing about them is that they are sound and complete. Now, what *I* meant was that if in the usual set of three rules you replace the augmentation rule with the other rule then your inference mechanisme will still be sound (the rule is correct) but no longer complete (it is weaker). So in that sense the two rules are not equivalent: in the usual list of three rules you cannot simply replace one with the other without changing the behavior of the inference mechanism.

I really don't see what other meaningful interpretation of "the augmentation rule can also be stated as ..." there could be. Note by the way that in this sense the augmentation rule is equivalent with the other rule plus the union rule:

3) {X->Y, X->Z } |= X->YZ

• Jan Hidders