YES * Step 1: UnsatPaths YES + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A] (?,1) 2. f3(A) -> f11(A) [A >= 0 && A >= 10 && 0 >= 1 + B] (?,1) 3. f3(A) -> f11(A) [A >= 0 && A >= 10] (?,1) Signature: {(f0,1);(f11,1);(f3,1)} Flow Graph: [0->{1,2,3},1->{1,2,3},2->{},3->{}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(0,2),(0,3)] * Step 2: FromIts YES + Considered Problem: Rules: 0. f0(A) -> f3(0) True (1,1) 1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A] (?,1) 2. f3(A) -> f11(A) [A >= 0 && A >= 10 && 0 >= 1 + B] (?,1) 3. f3(A) -> f11(A) [A >= 0 && A >= 10] (?,1) Signature: {(f0,1);(f11,1);(f3,1)} Flow Graph: [0->{1},1->{1,2,3},2->{},3->{}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: f0(A) -> f3(0) True f3(A) -> f3(1 + A) [A >= 0 && 9 >= A] f3(A) -> f11(A) [A >= 0 && A >= 10 && 0 >= 1 + B] f3(A) -> f11(A) [A >= 0 && A >= 10] Signature: {(f0,1);(f11,1);(f3,1)} Rule Graph: [0->{1},1->{1,2,3},2->{},3->{}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3] | `- p:[1] c: [1] * Step 4: CloseWith YES + Considered Problem: (Rules: f0(A) -> f3(0) True f3(A) -> f3(1 + A) [A >= 0 && 9 >= A] f3(A) -> f11(A) [A >= 0 && A >= 10 && 0 >= 1 + B] f3(A) -> f11(A) [A >= 0 && A >= 10] Signature: {(f0,1);(f11,1);(f3,1)} Rule Graph: [0->{1},1->{1,2,3},2->{},3->{}] ,We construct a looptree: P: [0,1,2,3] | `- p:[1] c: [1]) + Applied Processor: CloseWith True + Details: () YES