YES * Step 1: FromIts YES + Considered Problem: Rules: 0. f(A,B,C) -> g(A,1,1) True (1,1) 1. g(A,B,C) -> g(-1 + A,2*B,C) [-1 + A >= 0] (?,1) 2. g(A,B,C) -> h(A,B,C) [0 >= A] (?,1) 3. h(A,B,C) -> h(A,-1 + B,2*C) [-1 + B >= 0] (?,1) 4. h(A,B,C) -> i(A,B,C) [0 >= B] (?,1) 5. i(A,B,C) -> i(A,B,-1 + C) [-1 + C >= 0] (?,1) Signature: {(f,3);(g,3);(h,3);(i,3)} Flow Graph: [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{5},5->{5}] + Applied Processor: FromIts + Details: () * Step 2: Decompose YES + Considered Problem: Rules: f(A,B,C) -> g(A,1,1) True g(A,B,C) -> g(-1 + A,2*B,C) [-1 + A >= 0] g(A,B,C) -> h(A,B,C) [0 >= A] h(A,B,C) -> h(A,-1 + B,2*C) [-1 + B >= 0] h(A,B,C) -> i(A,B,C) [0 >= B] i(A,B,C) -> i(A,B,-1 + C) [-1 + C >= 0] Signature: {(f,3);(g,3);(h,3);(i,3)} Rule Graph: [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{5},5->{5}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4,5] | +- p:[1] c: [1] | +- p:[3] c: [3] | `- p:[5] c: [5] * Step 3: CloseWith YES + Considered Problem: (Rules: f(A,B,C) -> g(A,1,1) True g(A,B,C) -> g(-1 + A,2*B,C) [-1 + A >= 0] g(A,B,C) -> h(A,B,C) [0 >= A] h(A,B,C) -> h(A,-1 + B,2*C) [-1 + B >= 0] h(A,B,C) -> i(A,B,C) [0 >= B] i(A,B,C) -> i(A,B,-1 + C) [-1 + C >= 0] Signature: {(f,3);(g,3);(h,3);(i,3)} Rule Graph: [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{5},5->{5}] ,We construct a looptree: P: [0,1,2,3,4,5] | +- p:[1] c: [1] | +- p:[3] c: [3] | `- p:[5] c: [5]) + Applied Processor: CloseWith True + Details: () YES