YES * Step 1: FromIts YES + Considered Problem: Rules: 0. f2(A,B) -> f2(-1 + A,B) [B >= 0 && A >= 1] (?,1) 1. f0(A,B) -> f2(C,-1 + B) [B >= 1] (1,1) 2. f2(A,B) -> f2(C,-1 + B) [B >= 0 && B >= 1 && 0 >= A] (?,1) Signature: {(f0,2);(f2,2)} Flow Graph: [0->{0,2},1->{0,2},2->{0,2}] + Applied Processor: FromIts + Details: () * Step 2: Decompose YES + Considered Problem: Rules: f2(A,B) -> f2(-1 + A,B) [B >= 0 && A >= 1] f0(A,B) -> f2(C,-1 + B) [B >= 1] f2(A,B) -> f2(C,-1 + B) [B >= 0 && B >= 1 && 0 >= A] Signature: {(f0,2);(f2,2)} Rule Graph: [0->{0,2},1->{0,2},2->{0,2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2] | `- p:[0,2] c: [2] | `- p:[0] c: [0] * Step 3: CloseWith YES + Considered Problem: (Rules: f2(A,B) -> f2(-1 + A,B) [B >= 0 && A >= 1] f0(A,B) -> f2(C,-1 + B) [B >= 1] f2(A,B) -> f2(C,-1 + B) [B >= 0 && B >= 1 && 0 >= A] Signature: {(f0,2);(f2,2)} Rule Graph: [0->{0,2},1->{0,2},2->{0,2}] ,We construct a looptree: P: [0,1,2] | `- p:[0,2] c: [2] | `- p:[0] c: [0]) + Applied Processor: CloseWith True + Details: () YES