Hi John, I think I understand your point. The solution should be another inductive function - Flow s(@(c) b) = c. - Flow s(@(c) (1, v = e)) = c. - Flow s(@(c1) (@(c2) S)) = c2. - Flow s(@(c) (S1 ##1 S2)) = Flow(@(Flow(@(c)S1))S2). - Flow s(@(c) (S1 ##0 S2)) = Flow(@(Flow(@(c)S1))S2). - Flow s(@(c) (S1 or S2)) = if Flow(@(c)S1)=Flow(@(c)s2) then Flow(@(c)S1) otherwise 0 . - Flow s(@(c) (S1 intersect S2)) = if Flow(@(c)S1)=Flow(@(c)s2) then Flow(@(c)S1) otherwise 0. - Flow s(@(c) (first match S)) = Flow(@(c)S). - Flow s(@(c) (S[*0])) = c. - Flow s(@(c) (S[*1:$])) = Flow(@(c)S). Then, - T p(@(c) (S |-> Q) = (T s(@(c) S) |-> T p(@(Flow(@c(S)) Q)). With the restriction that Flow(@c(S)) is the only leading clock of Q (either because of inheritance or explicitly set) Doron -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.Received on Wed Sep 5 05:27:11 2007
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