YES * Step 1: UnsatRules YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,-1 + H,L) [1 >= A && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 1. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 2. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 3. lbl71(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,1,E,F,G,H,1 + I,J,K,L) [D >= 0 && I >= D && K >= 1 + I && A >= 1 + K && H = A] (?,1) 4. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [B >= 0 && A >= 2 + B && I = 1 + B && D = 0 && K = 1 + B && H = A] (?,1) 5. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [0 >= 1 + D && B >= 0 && D >= 0 && 1 >= D && A >= 2 + B && I = 1 + B && K = 1 + B && H = A] (?,1) 6. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [B >= 0 && A >= 2 + B && D = 1 && I = 1 + B && K = 1 + B && H = A] (?,1) 7. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,D,E,M,G,H,I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 8. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,D,E,F,G,H,1 + I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 9. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [A >= 2 && K = 0 && D = 1 && H = A && I = 1 && B = 0] (?,1) 10. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 11. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 12. start0(A,B,C,D,E,F,G,H,I,J,K,L) -> start(A,C,C,E,E,G,G,A,J,J,L,L) True (1,1) Signature: {(lbl13,12);(lbl53,12);(lbl71,12);(start,12);(start0,12);(stop,12)} Flow Graph: [0->{},1->{3},2->{4,5,6,7,8},3->{4,5,6,7,8},4->{},5->{9,10,11},6->{9,10,11},7->{3},8->{4,5,6,7,8},9->{} ,10->{3},11->{4,5,6,7,8},12->{0,1,2}] + Applied Processor: UnsatRules + Details: Following transitions have unsatisfiable constraints and are removed: [5] * Step 2: UnsatPaths YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,-1 + H,L) [1 >= A && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 1. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 2. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 3. lbl71(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,1,E,F,G,H,1 + I,J,K,L) [D >= 0 && I >= D && K >= 1 + I && A >= 1 + K && H = A] (?,1) 4. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [B >= 0 && A >= 2 + B && I = 1 + B && D = 0 && K = 1 + B && H = A] (?,1) 6. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [B >= 0 && A >= 2 + B && D = 1 && I = 1 + B && K = 1 + B && H = A] (?,1) 7. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,D,E,M,G,H,I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 8. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,D,E,F,G,H,1 + I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 9. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [A >= 2 && K = 0 && D = 1 && H = A && I = 1 && B = 0] (?,1) 10. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 11. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 12. start0(A,B,C,D,E,F,G,H,I,J,K,L) -> start(A,C,C,E,E,G,G,A,J,J,L,L) True (1,1) Signature: {(lbl13,12);(lbl53,12);(lbl71,12);(start,12);(start0,12);(stop,12)} Flow Graph: [0->{},1->{3},2->{4,6,7,8},3->{4,6,7,8},4->{},6->{9,10,11},7->{3},8->{4,6,7,8},9->{},10->{3},11->{4,6,7,8} ,12->{0,1,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(2,6),(3,4),(11,6)] * Step 3: FromIts YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,-1 + H,L) [1 >= A && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 1. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 2. start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] (?,1) 3. lbl71(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,1,E,F,G,H,1 + I,J,K,L) [D >= 0 && I >= D && K >= 1 + I && A >= 1 + K && H = A] (?,1) 4. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [B >= 0 && A >= 2 + B && I = 1 + B && D = 0 && K = 1 + B && H = A] (?,1) 6. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [B >= 0 && A >= 2 + B && D = 1 && I = 1 + B && K = 1 + B && H = A] (?,1) 7. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,D,E,M,G,H,I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 8. lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,D,E,F,G,H,1 + I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] (?,1) 9. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [A >= 2 && K = 0 && D = 1 && H = A && I = 1 && B = 0] (?,1) 10. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 11. lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] (?,1) 12. start0(A,B,C,D,E,F,G,H,I,J,K,L) -> start(A,C,C,E,E,G,G,A,J,J,L,L) True (1,1) Signature: {(lbl13,12);(lbl53,12);(lbl71,12);(start,12);(start0,12);(stop,12)} Flow Graph: [0->{},1->{3},2->{4,7,8},3->{6,7,8},4->{},6->{9,10,11},7->{3},8->{4,6,7,8},9->{},10->{3},11->{4,7,8} ,12->{0,1,2}] + Applied Processor: FromIts + Details: () * Step 4: Decompose YES + Considered Problem: Rules: start(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,-1 + H,L) [1 >= A && B = C && D = E && F = G && H = A && I = J && K = L] start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] lbl71(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,1,E,F,G,H,1 + I,J,K,L) [D >= 0 && I >= D && K >= 1 + I && A >= 1 + K && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [B >= 0 && A >= 2 + B && I = 1 + B && D = 0 && K = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [B >= 0 && A >= 2 + B && D = 1 && I = 1 + B && K = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,D,E,M,G,H,I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,D,E,F,G,H,1 + I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [A >= 2 && K = 0 && D = 1 && H = A && I = 1 && B = 0] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] start0(A,B,C,D,E,F,G,H,I,J,K,L) -> start(A,C,C,E,E,G,G,A,J,J,L,L) True Signature: {(lbl13,12);(lbl53,12);(lbl71,12);(start,12);(start0,12);(stop,12)} Rule Graph: [0->{},1->{3},2->{4,7,8},3->{6,7,8},4->{},6->{9,10,11},7->{3},8->{4,6,7,8},9->{},10->{3},11->{4,7,8} ,12->{0,1,2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4,6,7,8,9,10,11,12] | `- p:[3,7,8,11,6,10] c: [11] | `- p:[3,7,8,10,6] c: [6,10] | `- p:[3,7,8] c: [8] | `- p:[3,7] c: [3,7] * Step 5: CloseWith YES + Considered Problem: (Rules: start(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,-1 + H,L) [1 >= A && B = C && D = E && F = G && H = A && I = J && K = L] start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] start(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,-1 + H,L) [A >= 2 && B = C && D = E && F = G && H = A && I = J && K = L] lbl71(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,1,E,F,G,H,1 + I,J,K,L) [D >= 0 && I >= D && K >= 1 + I && A >= 1 + K && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [B >= 0 && A >= 2 + B && I = 1 + B && D = 0 && K = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl13(A,B,C,D,E,F,G,H,I,J,-1 + K,L) [B >= 0 && A >= 2 + B && D = 1 && I = 1 + B && K = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,D,E,M,G,H,I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,I,C,D,E,F,G,H,1 + I,J,K,L) [K >= 2 + B && B >= 0 && D >= 0 && K >= 1 + B && 1 >= D && A >= 1 + K && I = 1 + B && H = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> stop(A,B,C,D,E,F,G,H,I,J,K,L) [A >= 2 && K = 0 && D = 1 && H = A && I = 1 && B = 0] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl71(A,B,C,0,E,M,G,H,0,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] lbl13(A,B,C,D,E,F,G,H,I,J,K,L) -> lbl53(A,0,C,0,E,F,G,H,1,J,K,L) [B >= 1 && B >= 0 && A >= 2 + B && D = 1 && H = A && K = B && I = 1 + B] start0(A,B,C,D,E,F,G,H,I,J,K,L) -> start(A,C,C,E,E,G,G,A,J,J,L,L) True Signature: {(lbl13,12);(lbl53,12);(lbl71,12);(start,12);(start0,12);(stop,12)} Rule Graph: [0->{},1->{3},2->{4,7,8},3->{6,7,8},4->{},6->{9,10,11},7->{3},8->{4,6,7,8},9->{},10->{3},11->{4,7,8} ,12->{0,1,2}] ,We construct a looptree: P: [0,1,2,3,4,6,7,8,9,10,11,12] | `- p:[3,7,8,11,6,10] c: [11] | `- p:[3,7,8,10,6] c: [6,10] | `- p:[3,7,8] c: [8] | `- p:[3,7] c: [3,7]) + Applied Processor: CloseWith True + Details: () YES