YES(?,O(1)) * Step 1: UnsatPaths WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f3(A,B,C,D) -> f0(0,B,C,D) True (1,1) 1. f0(A,B,C,D) -> f0(1 + A,B,A,D) [A >= 0 && 9 >= A] (?,1) 2. f0(A,B,C,D) -> f4(A,E,C,D) [A >= 0 && A >= 10] (?,1) Signature: {(f0,4);(f3,4);(f4,4)} 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. f3(A,B,C,D) -> f0(0,B,C,D) True (1,1) 1. f0(A,B,C,D) -> f0(1 + A,B,A,D) [A >= 0 && 9 >= A] (?,1) 2. f0(A,B,C,D) -> f4(A,E,C,D) [A >= 0 && A >= 10] (?,1) Signature: {(f0,4);(f3,4);(f4,4)} 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. f3(A,B,C,D) -> f0(0,B,C,D) True (1,1) 1. f0(A,B,C,D) -> f0(1 + A,B,A,D) [A >= 0 && 9 >= A] (?,1) 2. f0(A,B,C,D) -> f4(A,E,C,D) [A >= 0 && A >= 10] (1,1) Signature: {(f0,4);(f3,4);(f4,4)} 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 + -1*x1 p(f3) = 10 p(f4) = 10 + -1*x1 Following rules are strictly oriented: [A >= 0 && 9 >= A] ==> f0(A,B,C,D) = 10 + -1*A > 9 + -1*A = f0(1 + A,B,A,D) Following rules are weakly oriented: True ==> f3(A,B,C,D) = 10 >= 10 = f0(0,B,C,D) [A >= 0 && A >= 10] ==> f0(A,B,C,D) = 10 + -1*A >= 10 + -1*A = f4(A,E,C,D) * Step 4: KnowledgePropagation WORST_CASE(?,O(1)) + Considered Problem: Rules: 0. f3(A,B,C,D) -> f0(0,B,C,D) True (1,1) 1. f0(A,B,C,D) -> f0(1 + A,B,A,D) [A >= 0 && 9 >= A] (10,1) 2. f0(A,B,C,D) -> f4(A,E,C,D) [A >= 0 && A >= 10] (1,1) Signature: {(f0,4);(f3,4);(f4,4)} Flow Graph: [0->{1},1->{1,2},2->{}] + Applied Processor: KnowledgePropagation + Details: The problem is already solved. YES(?,O(1))