YES(?,O(1)) * Step 1: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B) -> f6(C,0) True (1,1) 1. f6(A,B) -> f6(A,1 + B) [B >= 0 && 9 >= B] (?,1) 2. f6(A,B) -> f15(A,B) [B >= 0 && B >= 10] (?,1) Signature: {(f0,2);(f15,2);(f6,2)} Flow Graph: [0->{1,2},1->{1,2},2->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,2)] * Step 2: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B) -> f6(C,0) True (1,1) 1. f6(A,B) -> f6(A,1 + B) [B >= 0 && 9 >= B] (?,1) 2. f6(A,B) -> f15(A,B) [B >= 0 && B >= 10] (?,1) Signature: {(f0,2);(f15,2);(f6,2)} Flow Graph: [0->{1},1->{1,2},2->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B) -> f6(C,0) True (1,1) 1. f6(A,B) -> f6(A,1 + B) [B >= 0 && 9 >= B] (?,1) 2. f6(A,B) -> f15(A,B) [B >= 0 && B >= 10] (1,1) Signature: {(f0,2);(f15,2);(f6,2)} Flow Graph: [0->{1},1->{1,2},2->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 10 p(f15) = 10 + -1*x2 p(f6) = 10 + -1*x2 Following rules are strictly oriented: [B >= 0 && 9 >= B] ==> f6(A,B) = 10 + -1*B > 9 + -1*B = f6(A,1 + B) Following rules are weakly oriented: True ==> f0(A,B) = 10 >= 10 = f6(C,0) [B >= 0 && B >= 10] ==> f6(A,B) = 10 + -1*B >= 10 + -1*B = f15(A,B) * Step 4: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B) -> f6(C,0) True (1,1) 1. f6(A,B) -> f6(A,1 + B) [B >= 0 && 9 >= B] (10,1) 2. f6(A,B) -> f15(A,B) [B >= 0 && B >= 10] (1,1) Signature: {(f0,2);(f15,2);(f6,2)} Flow Graph: [0->{1},1->{1,2},2->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))