. Proof, It is obvious that any node v ? V that does not satisfy properties (a)a n d (b) is enabled and when v executes rule [R] during r 0 , v will satisfy both of these properties because v.m new = null or v.m new ? N (v)andv.s new = ? or |v

. Proof, A node executing rule [R3] four times would execute rule [R6] at least three times, R4] at most 6?d(v) times, i.e. it changes from state In to state Wait at most 6?d(v) times

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