YES(?,O(n^1)) * Step 1: TrivialSCCs WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. l0(A,B) -> l1(0,B) True (1,1) 1. l1(A,B) -> l1(1 + A,-1 + B) [A >= 0 && B >= 1] (?,1) 2. l1(A,B) -> l2(A,B) [A >= 0 && 0 >= B] (?,1) Signature: {(l0,2);(l1,2);(l2,2)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: PolyRank WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. l0(A,B) -> l1(0,B) True (1,1) 1. l1(A,B) -> l1(1 + A,-1 + B) [A >= 0 && B >= 1] (?,1) 2. l1(A,B) -> l2(A,B) [A >= 0 && 0 >= B] (1,1) Signature: {(l0,2);(l1,2);(l2,2)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(l0) = x2 p(l1) = x2 p(l2) = x2 Following rules are strictly oriented: [A >= 0 && B >= 1] ==> l1(A,B) = B > -1 + B = l1(1 + A,-1 + B) Following rules are weakly oriented: True ==> l0(A,B) = B >= B = l1(0,B) [A >= 0 && 0 >= B] ==> l1(A,B) = B >= B = l2(A,B) * Step 3: KnowledgePropagation WORST_CASE(?,O(n^1)) + Considered Problem: Rules: 0. l0(A,B) -> l1(0,B) True (1,1) 1. l1(A,B) -> l1(1 + A,-1 + B) [A >= 0 && B >= 1] (B,1) 2. l1(A,B) -> l2(A,B) [A >= 0 && 0 >= B] (1,1) Signature: {(l0,2);(l1,2);(l2,2)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(n^1))