On to Hydro

[Dean Tribble]

Specifically:

a <= b  primitive
a >= b          b <= a
a == b          a <= b && b <= a
a != b          !(a == b)

For the operation 'mu' meaning not comparable (we didn't actually define 
such an operation :-)
a mu b          !(a <= b || b <= a)

Finally, my vague reconstruction of the "strict" operations:
a > b           b <= a && !(a <= b)
a < b           a <= b && !(b <= a)

Obviously the strict operations could implemented directly and efficiently 
for fully ordered spaces.

At 07:03 AM 8/18/2000, Jonathan S. Shapiro wrote:
>Yes. In a partial order, there is no guarantee that two random elements are
>comparable. The way to visualize this is to think of a lattice. Comparisons
>can only be performed between elements that are (transitively) connected in
>the lattice.
>
>Thus, both < and <= have different meanings than in a total order.
>
> > Put another way, does '<' have some different meaning in a world
> > with only partial orders?
> >
> > Tyler