At 12:47 30/07/04 +0200, I wrote: >Oh, any accurate machine (for which Bp->p is true) is obviously normal.
This is false. But an accurate stable machine will be stable. Just substitute p with Bp in (Bp -> p) to get BBp -> Bp. That's stability, not normality.
Bruno
http://iridia.ulb.ac.be/~marchal/
