Relating Computation and Time

Consider any action y 00 in this set. Since y 00 2 B 00 x , we have x 00 y 00. It now remains to be shown that y 00 2 Before(y). Suppose y 00 6 2 Before(y). Then x 0 y 00 (because x 0 y 0). Consequently, y 00 2 B 0 x which also contradicts the selection of y 00. So y 00 2 Before(y). if: Consider instants t = (A; B) and t 0 = (A 0 ; B 0) such that t < t 0. We show that Suppose that the condition does not hold. Then for any x in A and any y in (B ? B 0), since y 2 B, x y. Now consider any x 0 in After(A). Since y 2 Before(B 0), from the negation of condition 5 we have :(x 0 y). Since, for any y 0 2 B 0 , x 0 y 0 , the actions x; y; x 0 ; y 0 satisfy the premise of the condition of the theorem. Hence, there exist x 00 ; y 00 2 ACT such that x 00 y 00 ^ x 00 2 After(x) ^ y 00 2 Before(y 0) But, y 0 is an arbitrary element of B 0. Therefore, since y 00 2 Before(y 0), such a y 00 can be selected from Before(B 0), and condition 5 holds. Thus the negation of condition 5 leads to a contradiction. Therefore, condition 5 holds. Now consider the instant x 00 + = (A 00 x ; B 00 x). Since x 00 2 A 00 x and x 00 6 2 A, we have t < x 00 +. Also, y 00 2 Before(B 0). Thus, y 00 6 2 B 0 and, consequently, x 00 + < t 0. So an instant x 00 + exists such that t < x 00 + < t 0. 2 16 5] L. Lamport. Time, clocks, and the ordering of events in a distributed system.