On 10 November 2011 21:32, Lachlan Andrew <[email protected]> wrote: > The proof of Theorem 1 seems to assume that > (("for all queues, P(S)=1 implies P(B)=0" is false) > and (for all queues, P(B)=0 or P(B)=1)) > implies > ("for all queues, P(S)=1 implies P(B)=1" is true). > That is not the case. It simply implies > (there exists a queue such that either P(B)=1 or P(S)\neq 1)
Oops. Before someone else points it out: not only did I get deMorgan's law wrong, I see that my initial interpretation of your argument was wrong. I'll keep looking to see where the flaw is. L -- Lachlan Andrew Centre for Advanced Internet Architectures (CAIA) Swinburne University of Technology, Melbourne, Australia <http://caia.swin.edu.au/cv/landrew> Ph +61 3 9214 4837 _______________________________________________ IEEE Communications Society Tech. Committee on Computer Communications (TCCC) - for discussions on computer networking and communication. [email protected] https://lists.cs.columbia.edu/cucslists/listinfo/tccc
