, To prove the claim, observe that ? reachesv at Tv(? v * ) + ? v * , so if we back-propagate Tv(? v * ) + ? v * fromv to v i alongp(i), we reach every intermediary node v j at C(?(j)) = T v j (? v * ) + ? v * . Using that ?v(? k ) ? Tv(? v * ) + ? v * , we observe that if we back-propagate ?v(? k ) fromv to v i alongp(i), we necessarily reach the intermediary nodes v j later than when back-propagating Tv(? v * ) + ? v * , i.e., later than T v j (? v * ) + ? v * . Moreover, if one intermediary node is reached later than ? v j (? k ), then Algorithm 2 necessarily stops at step 2, which is not possible if?v < +?. By T v j (? v * ) + ? v * ? T v j (? k ) + ? k , we get that every intermediary node v j is also reached in the time interval, conclude that?v ? t v i ?v ? T v i (? k ) ? ?v(? k ) ? Tv(? v * ) ? ? v * , where the last inequality follows from ?v(? k ) ? Tv(? v * ) + ? v *

, Combining Propositions 1 and equation (19) proves the validity of Algorithm 3

. Proof, Equation (19) shows that the sequence ? 1 , . . . , ? K is constructed as expected

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