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