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