YES * Step 1: TrivialSCCs YES + Considered Problem: Rules: 0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F) True (1,1) 1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1] (?,1) 2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1) 3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3] (?,1) 4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F) [0 >= A] (?,1) Signature: {(f0,6);(f1,6);(f2,6)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}] + Applied Processor: TrivialSCCs + Details: All trivial SCCs of the transition graph admit timebound 1. * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F) True (1,1) 1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1] (?,1) 2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1) 3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3] (?,1) 4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F) [0 >= A] (1,1) Signature: {(f0,6);(f1,6);(f2,6)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,4)] * Step 3: Looptree YES + Considered Problem: Rules: 0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F) True (1,1) 1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1] (?,1) 2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1) 3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3] (?,1) 4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F) [0 >= A] (1,1) Signature: {(f0,6);(f1,6);(f2,6)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}] + Applied Processor: Looptree + Details: We construct a looptree: P: [0,1,2,3,4] | `- p:[1,2,3] c: [2] | `- p:[1,3] c: [1] | `- p:[3] c: [3] YES