Re: So let me get this right: (Was: NFNF vs 1NF ...)

Mikito Harakiri wrote:
>
> http://arxiv.org/abs/cs.DB/0501053
> demonstrates that relational algebra could be formulated in terms of only
> two operators which are based upon equality. This can't be achieved without
> admitting certain infinite relations.

Very interesting link. But you not only have to admit them as intermediate results, you need them in your base language in order to express certain queries. Strictly speaking the language consists of the two operators plus an (infinite?) set of these infinite relations. I'm not sure if that simplifies much.

> It's hard to tell how useful that 2
> operation algebra is (especially as it's non-distributive), but at least
> we've got some resemblance to the "established" algebras in the math field.
> For one thing, solving equations in 6-operation algebra is a proposition for
> extreme adventurers only, while equations with only 2 operations seem
> reasonable.

Strictly speaking Vadim doesn't really prove that because he only shows that he can simulate the manipulation of binary relations, which is strictly weaker than full first order logic. But it could be that his proof can be generalized for the n-ary case, I haven't checked that.

• Jan Hidders