YES * Step 1: UnsatPaths YES + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (?,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3,4},4->{2,3,4}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,4),(4,3)] * Step 2: FromIts YES + Considered Problem: Rules: 0. f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] (1,1) 1. f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] (1,1) 2. f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] (?,1) 3. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] (?,1) 4. f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] (?,1) Signature: {(f1,5);(f2,5);(f300,5)} Flow Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] Signature: {(f1,5);(f2,5);(f300,5)} Rule Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4] | +- p:[4] c: [4] | `- p:[3] c: [3] * Step 4: CloseWith YES + Considered Problem: (Rules: f2(A,B,C,D,E) -> f1(A,F,C,D,E) [1 >= A] f2(A,B,C,D,E) -> f300(A,B,C,D,E) [A >= 2 && C >= 2] f300(A,B,C,D,E) -> f1(A,F,C,D,E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 1 + D >= 0] f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && E >= 0 && 0 >= 2 + D] f300(A,B,C,D,E) -> f300(A,B,C,1 + D,1 + E) [-2 + C >= 0 && -4 + A + C >= 0 && -2 + A >= 0 && 0 >= 2 + E && 0 >= 2 + D] Signature: {(f1,5);(f2,5);(f300,5)} Rule Graph: [0->{},1->{2,3,4},2->{},3->{2,3},4->{2,4}] ,We construct a looptree: P: [0,1,2,3,4] | +- p:[4] c: [4] | `- p:[3] c: [3]) + Applied Processor: CloseWith True + Details: () YES