YES * Step 1: UnsatPaths YES + Considered Problem: Rules: 0. start(A,B) -> eval(A,B) True (1,1) 1. eval(A,B) -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B] (?,1) 2. eval(A,B) -> eval(A,0) [A >= 1 && B = 1] (?,1) 3. eval(A,B) -> eval(A,A) [B >= 1 && 1 + B >= 0 && B >= 1 + A] (?,1) 4. eval(A,B) -> eval(A,A) [0 >= A && B = 1] (?,1) Signature: {(eval,2);(start,2)} Flow Graph: [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{1,2,3,4}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(1,3) ,(1,4) ,(2,1) ,(2,2) ,(2,3) ,(2,4) ,(3,3) ,(3,4) ,(4,1) ,(4,2) ,(4,3) ,(4,4)] * Step 2: FromIts YES + Considered Problem: Rules: 0. start(A,B) -> eval(A,B) True (1,1) 1. eval(A,B) -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B] (?,1) 2. eval(A,B) -> eval(A,0) [A >= 1 && B = 1] (?,1) 3. eval(A,B) -> eval(A,A) [B >= 1 && 1 + B >= 0 && B >= 1 + A] (?,1) 4. eval(A,B) -> eval(A,A) [0 >= A && B = 1] (?,1) Signature: {(eval,2);(start,2)} Flow Graph: [0->{1,2,3,4},1->{1,2},2->{},3->{1,2},4->{}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: start(A,B) -> eval(A,B) True eval(A,B) -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B] eval(A,B) -> eval(A,0) [A >= 1 && B = 1] eval(A,B) -> eval(A,A) [B >= 1 && 1 + B >= 0 && B >= 1 + A] eval(A,B) -> eval(A,A) [0 >= A && B = 1] Signature: {(eval,2);(start,2)} Rule Graph: [0->{1,2,3,4},1->{1,2},2->{},3->{1,2},4->{}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4] | `- p:[1] c: [1] * Step 4: CloseWith YES + Considered Problem: (Rules: start(A,B) -> eval(A,B) True eval(A,B) -> eval(A,-1 + B) [B >= 1 && 1 + B >= 0 && A >= B] eval(A,B) -> eval(A,0) [A >= 1 && B = 1] eval(A,B) -> eval(A,A) [B >= 1 && 1 + B >= 0 && B >= 1 + A] eval(A,B) -> eval(A,A) [0 >= A && B = 1] Signature: {(eval,2);(start,2)} Rule Graph: [0->{1,2,3,4},1->{1,2},2->{},3->{1,2},4->{}] ,We construct a looptree: P: [0,1,2,3,4] | `- p:[1] c: [1]) + Applied Processor: CloseWith True + Details: () YES