YES * Step 1: UnsatPaths YES + Considered Problem: Rules: 0. f0(A,B) -> f4(0,B) True (1,1) 1. f4(A,B) -> f4(1 + A,B) [1 >= A] (?,1) 2. f10(A,B) -> f10(A,1 + B) [1 >= B] (?,1) 3. f10(A,B) -> f18(A,B) [B >= 2 && 0 >= 1 + C] (?,1) 4. f10(A,B) -> f18(A,B) [B >= 2] (?,1) 5. f4(A,B) -> f10(A,0) [A >= 2] (?,1) Signature: {(f0,2);(f10,2);(f18,2);(f4,2)} Flow Graph: [0->{1,5},1->{1,5},2->{2,3,4},3->{},4->{},5->{2,3,4}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,5),(5,3),(5,4)] * Step 2: FromIts YES + Considered Problem: Rules: 0. f0(A,B) -> f4(0,B) True (1,1) 1. f4(A,B) -> f4(1 + A,B) [1 >= A] (?,1) 2. f10(A,B) -> f10(A,1 + B) [1 >= B] (?,1) 3. f10(A,B) -> f18(A,B) [B >= 2 && 0 >= 1 + C] (?,1) 4. f10(A,B) -> f18(A,B) [B >= 2] (?,1) 5. f4(A,B) -> f10(A,0) [A >= 2] (?,1) Signature: {(f0,2);(f10,2);(f18,2);(f4,2)} Flow Graph: [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: f0(A,B) -> f4(0,B) True f4(A,B) -> f4(1 + A,B) [1 >= A] f10(A,B) -> f10(A,1 + B) [1 >= B] f10(A,B) -> f18(A,B) [B >= 2 && 0 >= 1 + C] f10(A,B) -> f18(A,B) [B >= 2] f4(A,B) -> f10(A,0) [A >= 2] Signature: {(f0,2);(f10,2);(f18,2);(f4,2)} Rule Graph: [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4,5] | +- p:[1] c: [1] | `- p:[2] c: [2] * Step 4: CloseWith YES + Considered Problem: (Rules: f0(A,B) -> f4(0,B) True f4(A,B) -> f4(1 + A,B) [1 >= A] f10(A,B) -> f10(A,1 + B) [1 >= B] f10(A,B) -> f18(A,B) [B >= 2 && 0 >= 1 + C] f10(A,B) -> f18(A,B) [B >= 2] f4(A,B) -> f10(A,0) [A >= 2] Signature: {(f0,2);(f10,2);(f18,2);(f4,2)} Rule Graph: [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}] ,We construct a looptree: P: [0,1,2,3,4,5] | +- p:[1] c: [1] | `- p:[2] c: [2]) + Applied Processor: CloseWith True + Details: () YES