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