YES * Step 1: UnsatRules YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [0 >= A && B = C && D = A && E = F && G = H] (?,1) 1. start(A,B,C,D,E,F,G,H) -> lbl6(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && B = C && D = A && E = F && G = H] (?,1) 2. start(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1,F,0,H) [A >= 1 && C >= 1 + A && B = C && D = A && E = F && G = H] (?,1) 3. lbl6(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && G = H && E = F && D = A && B = C] (?,1) 4. lbl121(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [G >= C && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] (?,1) 5. lbl121(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 + E && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] (?,1) 6. lbl121(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [C >= 1 + G && A + C >= 3 + G && A >= 1 && G >= 0 && C >= 1 + A && E = A && D = A && B = C] (?,1) 7. lbl141(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [A >= 2 && C >= 1 + A && C >= 1 && G = C && E = 0 && D = A && B = C] (?,1) 8. lbl141(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] (?,1) 9. lbl141(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [0 >= A && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [C >= 1 + A && A >= 2 && E = 0 && G = C && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F,G,H) -> start(A,C,C,A,F,F,H,H) True (1,1) Signature: {(cut,8);(lbl121,8);(lbl141,8);(lbl6,8);(start,8);(start0,8);(stop,8)} Flow Graph: [0->{},1->{3},2->{4,5,6},3->{},4->{10},5->{4,5,6},6->{7,8,9},7->{10},8->{4,5,6},9->{7,8,9},10->{},11->{0,1 ,2}] + Applied Processor: UnsatRules + Details: Following transitions have unsatisfiable constraints and are removed: [4,9] * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [0 >= A && B = C && D = A && E = F && G = H] (?,1) 1. start(A,B,C,D,E,F,G,H) -> lbl6(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && B = C && D = A && E = F && G = H] (?,1) 2. start(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1,F,0,H) [A >= 1 && C >= 1 + A && B = C && D = A && E = F && G = H] (?,1) 3. lbl6(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && G = H && E = F && D = A && B = C] (?,1) 5. lbl121(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 + E && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] (?,1) 6. lbl121(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [C >= 1 + G && A + C >= 3 + G && A >= 1 && G >= 0 && C >= 1 + A && E = A && D = A && B = C] (?,1) 7. lbl141(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [A >= 2 && C >= 1 + A && C >= 1 && G = C && E = 0 && D = A && B = C] (?,1) 8. lbl141(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [C >= 1 + A && A >= 2 && E = 0 && G = C && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F,G,H) -> start(A,C,C,A,F,F,H,H) True (1,1) Signature: {(cut,8);(lbl121,8);(lbl141,8);(lbl6,8);(start,8);(start0,8);(stop,8)} Flow Graph: [0->{},1->{3},2->{5,6},3->{},5->{5,6},6->{7,8},7->{10},8->{5,6},10->{},11->{0,1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(8,6)] * Step 3: FromIts YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [0 >= A && B = C && D = A && E = F && G = H] (?,1) 1. start(A,B,C,D,E,F,G,H) -> lbl6(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && B = C && D = A && E = F && G = H] (?,1) 2. start(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1,F,0,H) [A >= 1 && C >= 1 + A && B = C && D = A && E = F && G = H] (?,1) 3. lbl6(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && G = H && E = F && D = A && B = C] (?,1) 5. lbl121(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 + E && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] (?,1) 6. lbl121(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [C >= 1 + G && A + C >= 3 + G && A >= 1 && G >= 0 && C >= 1 + A && E = A && D = A && B = C] (?,1) 7. lbl141(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [A >= 2 && C >= 1 + A && C >= 1 && G = C && E = 0 && D = A && B = C] (?,1) 8. lbl141(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] (?,1) 10. cut(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [C >= 1 + A && A >= 2 && E = 0 && G = C && D = A && B = C] (?,1) 11. start0(A,B,C,D,E,F,G,H) -> start(A,C,C,A,F,F,H,H) True (1,1) Signature: {(cut,8);(lbl121,8);(lbl141,8);(lbl6,8);(start,8);(start0,8);(stop,8)} Flow Graph: [0->{},1->{3},2->{5,6},3->{},5->{5,6},6->{7,8},7->{10},8->{5},10->{},11->{0,1,2}] + Applied Processor: FromIts + Details: () * Step 4: Decompose YES + Considered Problem: Rules: start(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [0 >= A && B = C && D = A && E = F && G = H] start(A,B,C,D,E,F,G,H) -> lbl6(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && B = C && D = A && E = F && G = H] start(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1,F,0,H) [A >= 1 && C >= 1 + A && B = C && D = A && E = F && G = H] lbl6(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && G = H && E = F && D = A && B = C] lbl121(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 + E && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] lbl121(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [C >= 1 + G && A + C >= 3 + G && A >= 1 && G >= 0 && C >= 1 + A && E = A && D = A && B = C] lbl141(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [A >= 2 && C >= 1 + A && C >= 1 && G = C && E = 0 && D = A && B = C] lbl141(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] cut(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [C >= 1 + A && A >= 2 && E = 0 && G = C && D = A && B = C] start0(A,B,C,D,E,F,G,H) -> start(A,C,C,A,F,F,H,H) True Signature: {(cut,8);(lbl121,8);(lbl141,8);(lbl6,8);(start,8);(start0,8);(stop,8)} Rule Graph: [0->{},1->{3},2->{5,6},3->{},5->{5,6},6->{7,8},7->{10},8->{5},10->{},11->{0,1,2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,5,6,7,8,10,11] | `- p:[5,8,6] c: [6,8] | `- p:[5] c: [5] * Step 5: CloseWith YES + Considered Problem: (Rules: start(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [0 >= A && B = C && D = A && E = F && G = H] start(A,B,C,D,E,F,G,H) -> lbl6(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && B = C && D = A && E = F && G = H] start(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1,F,0,H) [A >= 1 && C >= 1 + A && B = C && D = A && E = F && G = H] lbl6(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [A >= 1 && A >= C && G = H && E = F && D = A && B = C] lbl121(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 + E && C >= 1 + G && A + C >= 3 + G && A >= E && E >= 1 && G >= 0 && C >= 1 + A && D = A && B = C] lbl121(A,B,C,D,E,F,G,H) -> lbl141(A,B,C,D,0,F,1 + G,H) [C >= 1 + G && A + C >= 3 + G && A >= 1 && G >= 0 && C >= 1 + A && E = A && D = A && B = C] lbl141(A,B,C,D,E,F,G,H) -> cut(A,B,C,D,E,F,G,H) [A >= 2 && C >= 1 + A && C >= 1 && G = C && E = 0 && D = A && B = C] lbl141(A,B,C,D,E,F,G,H) -> lbl121(A,B,C,D,1 + E,F,G,H) [A >= 1 && C >= 1 + G && A >= 2 && C >= 1 + A && G >= 1 && C >= G && E = 0 && D = A && B = C] cut(A,B,C,D,E,F,G,H) -> stop(A,B,C,D,E,F,G,H) [C >= 1 + A && A >= 2 && E = 0 && G = C && D = A && B = C] start0(A,B,C,D,E,F,G,H) -> start(A,C,C,A,F,F,H,H) True Signature: {(cut,8);(lbl121,8);(lbl141,8);(lbl6,8);(start,8);(start0,8);(stop,8)} Rule Graph: [0->{},1->{3},2->{5,6},3->{},5->{5,6},6->{7,8},7->{10},8->{5},10->{},11->{0,1,2}] ,We construct a looptree: P: [0,1,2,3,5,6,7,8,10,11] | `- p:[5,8,6] c: [6,8] | `- p:[5] c: [5]) + Applied Processor: CloseWith True + Details: () YES