YES * Step 1: UnsatRules YES + Considered Problem: Rules: 0. eval(A,B) -> eval(-1 + A,B) [A >= 1] (?,1) 1. eval(A,B) -> eval(-1 + A,B) [B >= 1 && A >= 1] (?,1) 2. eval(A,B) -> eval(A,-1 + B) [A >= 1 && 0 >= A && B >= 1] (?,1) 3. eval(A,B) -> eval(A,-1 + B) [B >= 1 && 0 >= A] (?,1) 4. eval(A,B) -> eval(A,B) [A >= 1 && 0 >= A && 0 >= B] (?,1) 5. eval(A,B) -> eval(A,B) [B >= 1 && 0 >= A && 0 >= B] (?,1) 6. start(A,B) -> eval(A,B) True (1,1) Signature: {(eval,2);(start,2)} Flow Graph: [0->{0,1,2,3,4,5},1->{0,1,2,3,4,5},2->{0,1,2,3,4,5},3->{0,1,2,3,4,5},4->{0,1,2,3,4,5},5->{0,1,2,3,4,5} ,6->{0,1,2,3,4,5}] + Applied Processor: UnsatRules + Details: Following transitions have unsatisfiable constraints and are removed: [2,4,5] * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. eval(A,B) -> eval(-1 + A,B) [A >= 1] (?,1) 1. eval(A,B) -> eval(-1 + A,B) [B >= 1 && A >= 1] (?,1) 3. eval(A,B) -> eval(A,-1 + B) [B >= 1 && 0 >= A] (?,1) 6. start(A,B) -> eval(A,B) True (1,1) Signature: {(eval,2);(start,2)} Flow Graph: [0->{0,1,3},1->{0,1,3},3->{0,1,3},6->{0,1,3}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(3,0),(3,1)] * Step 3: FromIts YES + Considered Problem: Rules: 0. eval(A,B) -> eval(-1 + A,B) [A >= 1] (?,1) 1. eval(A,B) -> eval(-1 + A,B) [B >= 1 && A >= 1] (?,1) 3. eval(A,B) -> eval(A,-1 + B) [B >= 1 && 0 >= A] (?,1) 6. start(A,B) -> eval(A,B) True (1,1) Signature: {(eval,2);(start,2)} Flow Graph: [0->{0,1,3},1->{0,1,3},3->{3},6->{0,1,3}] + Applied Processor: FromIts + Details: () * Step 4: Decompose YES + Considered Problem: Rules: eval(A,B) -> eval(-1 + A,B) [A >= 1] eval(A,B) -> eval(-1 + A,B) [B >= 1 && A >= 1] eval(A,B) -> eval(A,-1 + B) [B >= 1 && 0 >= A] start(A,B) -> eval(A,B) True Signature: {(eval,2);(start,2)} Rule Graph: [0->{0,1,3},1->{0,1,3},3->{3},6->{0,1,3}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,3,6] | +- p:[0,1] c: [1] | | | `- p:[0] c: [0] | `- p:[3] c: [3] * Step 5: CloseWith YES + Considered Problem: (Rules: eval(A,B) -> eval(-1 + A,B) [A >= 1] eval(A,B) -> eval(-1 + A,B) [B >= 1 && A >= 1] eval(A,B) -> eval(A,-1 + B) [B >= 1 && 0 >= A] start(A,B) -> eval(A,B) True Signature: {(eval,2);(start,2)} Rule Graph: [0->{0,1,3},1->{0,1,3},3->{3},6->{0,1,3}] ,We construct a looptree: P: [0,1,3,6] | +- p:[0,1] c: [1] | | | `- p:[0] c: [0] | `- p:[3] c: [3]) + Applied Processor: CloseWith True + Details: () YES