YES * Step 1: FromIts YES + Considered Problem: Rules: 0. eval1(A,B) -> eval2(A,B) [A >= 1] (?,1) 1. eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] (?,1) 2. eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] (?,1) 3. start(A,B) -> eval1(A,B) True (1,1) Signature: {(eval1,2);(eval2,2);(start,2)} Flow Graph: [0->{1,2},1->{1,2},2->{0},3->{0}] + Applied Processor: FromIts + Details: () * Step 2: Decompose YES + Considered Problem: Rules: eval1(A,B) -> eval2(A,B) [A >= 1] eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] start(A,B) -> eval1(A,B) True Signature: {(eval1,2);(eval2,2);(start,2)} Rule Graph: [0->{1,2},1->{1,2},2->{0},3->{0}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3] | `- p:[0,2,1] c: [0,2] | `- p:[1] c: [1] * Step 3: CloseWith YES + Considered Problem: (Rules: eval1(A,B) -> eval2(A,B) [A >= 1] eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] start(A,B) -> eval1(A,B) True Signature: {(eval1,2);(eval2,2);(start,2)} Rule Graph: [0->{1,2},1->{1,2},2->{0},3->{0}] ,We construct a looptree: P: [0,1,2,3] | `- p:[0,2,1] c: [0,2] | `- p:[1] c: [1]) + Applied Processor: CloseWith True + Details: () YES