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