Re: Better management of mergejoinable operators

Andrew - Supernews <andrew+nonews@supernews.com> writes:
> On 2006-12-13, Tom Lane <tgl@sss.pgh.pa.us> wrote:
>> BTW, I think it's possible to prove that there need never be two for the
>> case of both sides the same datatype.

> Counterexample even for a single data type: define an operator x =* y
> which is true when 2x = y.  This is mergejoinable using the following
> operators: SORT1 = <, SORT2 = <, LTCMP = (2x < y), RTCMP = (2x > y)
> (which is of course the same sortops as for regular =).

I think not --- the corresponding sort operators would have to be
"2x < y" etc, else the trichotomy law fails, and so do all standard
sort algorithms.
        regards, tom lane