YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. f0(A,B,C) -> f8(D,0,C) True (1,1) 1. f8(A,B,C) -> f8(A,1 + B,C) [B >= 0 && 9 >= B] (?,1) 2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C] (?,1) 3. f19(A,B,C) -> f29(A,B,C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (?,1) 4. f8(A,B,C) -> f19(A,B,0) [B >= 0 && B >= 10] (?,1) Signature: {(f0,3);(f19,3);(f29,3);(f8,3)} Flow Graph: [0->{1,4},1->{1,4},2->{2,3},3->{},4->{2,3}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. f0(A,B,C) -> f8(D,0,C) True (1,1) 1. f8(A,B,C) -> f8(A,1 + B,C) [B >= 0 && 9 >= B] (?,1) 2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C] (?,1) 3. f19(A,B,C) -> f29(A,B,C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (1,1) 4. f8(A,B,C) -> f19(A,B,0) [B >= 0 && B >= 10] (1,1) Signature: {(f0,3);(f19,3);(f29,3);(f8,3)} Flow Graph: [0->{1,4},1->{1,4},2->{2,3},3->{},4->{2,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,4),(4,3)] * Step 3: Looptree YES + Considered Problem: Rules: 0. f0(A,B,C) -> f8(D,0,C) True (1,1) 1. f8(A,B,C) -> f8(A,1 + B,C) [B >= 0 && 9 >= B] (?,1) 2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C] (?,1) 3. f19(A,B,C) -> f29(A,B,C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (1,1) 4. f8(A,B,C) -> f19(A,B,0) [B >= 0 && B >= 10] (1,1) Signature: {(f0,3);(f19,3);(f29,3);(f8,3)} Flow Graph: [0->{1},1->{1,4},2->{2,3},3->{},4->{2}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4] | +- p:[1] c: [1] | `- p:[2] c: [2] YES