YES(?,O(1)) * Step 1: TrivialSCCs WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True (1,1) 1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f28(A,B,C,D,B,B,K,L,L,L) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] (?,1) Signature: {(f0,10);(f28,10);(f5,10)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True (1,1) 1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f28(A,B,C,D,B,B,K,L,L,L) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] (1,1) Signature: {(f0,10);(f28,10);(f5,10)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,3)] * Step 3: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True (1,1) 1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f28(A,B,C,D,B,B,K,L,L,L) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] (1,1) Signature: {(f0,10);(f28,10);(f5,10)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 32 p(f28) = 32 + -1*x3 p(f5) = 32 + -1*x3 Following rules are strictly oriented: [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C > 31 + -1*C = f5(A,B,1 + C,0,E,F,G,H,I,J) Following rules are weakly oriented: True ==> f0(A,B,C,D,E,F,G,H,I,J) = 32 >= 32 = f5(K,0,0,D,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C >= 31 + -1*C = f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C >= 32 + -1*C = f28(A,B,C,D,B,B,K,L,L,L) * Step 4: PolyRank WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True (1,1) 1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (?,1) 2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (32,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f28(A,B,C,D,B,B,K,L,L,L) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] (1,1) Signature: {(f0,10);(f28,10);(f5,10)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: PolyRank {useFarkas = True, withSizebounds = [], shape = Linear} + Details: We apply a polynomial interpretation of shape linear: p(f0) = 32 p(f28) = 32 + -1*x3 p(f5) = 32 + -1*x3 Following rules are strictly oriented: [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C > 31 + -1*C = f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C > 31 + -1*C = f5(A,B,1 + C,0,E,F,G,H,I,J) Following rules are weakly oriented: True ==> f0(A,B,C,D,E,F,G,H,I,J) = 32 >= 32 = f5(K,0,0,D,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] ==> f5(A,B,C,D,E,F,G,H,I,J) = 32 + -1*C >= 32 + -1*C = f28(A,B,C,D,B,B,K,L,L,L) * Step 5: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True (1,1) 1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (32,1) 2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 31 >= C] (32,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f28(A,B,C,D,B,B,K,L,L,L) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 32] (1,1) Signature: {(f0,10);(f28,10);(f5,10)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))