Doug Lea wrote:
> And further, if we had
> T1 T2 T3
> e; wx{x=1} rx{r1=x}; wy(y=1} ry{r2=y}
> we'd eventually get e --> ry
>
Oops. Not e --> ry directly. But it is contained in
T1's transitive closure, which is is required to be acyclic.
-Doug