YES * Step 1: UnsatPaths YES + 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) [15 >= 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) [15 >= C] (?,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L) [C >= 16] (?,1) Signature: {(f0,10);(f27,10);(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 2: FromIts YES + 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) [15 >= 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) [15 >= C] (?,1) 3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L) [C >= 16] (?,1) Signature: {(f0,10);(f27,10);(f5,10)} Flow Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [15 >= C] f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [15 >= C] f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L) [C >= 16] Signature: {(f0,10);(f27,10);(f5,10)} Rule Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3] | `- p:[1,2] c: [2] | `- p:[1] c: [1] * Step 4: CloseWith YES + Considered Problem: (Rules: f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J) True f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [15 >= C] f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J) [15 >= C] f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L) [C >= 16] Signature: {(f0,10);(f27,10);(f5,10)} Rule Graph: [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}] ,We construct a looptree: P: [0,1,2,3] | `- p:[1,2] c: [2] | `- p:[1] c: [1]) + Applied Processor: CloseWith True + Details: () YES