YES * Step 1: UnsatPaths YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,0,M,N) [1 >= A && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 1. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,0,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 2. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,1,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 3. lbl91(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl13(A,L,C,D,E,F,G,H,I,J,K,1 + L,M,N) [D >= H && D >= 1 && 1 + H >= D && A >= 1 + D && N = A && 1 + L = D && J = A] (?,1) 4. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl91(A,B,C,D,E,O,G,H,I,J,K,L,M,N) [D >= H && D >= 1 && A >= 1 + D && 1 + H >= D && J = A && 1 + L = D && N = A] (?,1) 5. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,H,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] (?,1) 6. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,J,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] (?,1) 7. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,L,M,N) [2 + H >= A && A >= 2 && A >= 1 + H && 1 + L = A && 2 + B = A && 1 + D = A && N = A && J = A] (?,1) 8. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] (?,1) 9. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,1 + L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] (?,1) 10. start0(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> start(A,C,C,E,E,G,G,I,I,K,K,M,M,A) True (1,1) Signature: {(lbl13,14);(lbl53,14);(lbl91,14);(start,14);(start0,14);(stop,14)} Flow Graph: [0->{},1->{4,5,6},2->{4,5,6},3->{7,8,9},4->{3},5->{4,5,6},6->{4,5,6},7->{},8->{4,5,6},9->{4,5,6},10->{0,1 ,2}] + Applied Processor: UnsatPaths + Details: We remove following edges from the transition graph: [(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)] * Step 2: FromIts YES + Considered Problem: Rules: 0. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,0,M,N) [1 >= A && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 1. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,0,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 2. start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,1,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] (?,1) 3. lbl91(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl13(A,L,C,D,E,F,G,H,I,J,K,1 + L,M,N) [D >= H && D >= 1 && 1 + H >= D && A >= 1 + D && N = A && 1 + L = D && J = A] (?,1) 4. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl91(A,B,C,D,E,O,G,H,I,J,K,L,M,N) [D >= H && D >= 1 && A >= 1 + D && 1 + H >= D && J = A && 1 + L = D && N = A] (?,1) 5. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,H,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] (?,1) 6. lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,J,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] (?,1) 7. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,L,M,N) [2 + H >= A && A >= 2 && A >= 1 + H && 1 + L = A && 2 + B = A && 1 + D = A && N = A && J = A] (?,1) 8. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] (?,1) 9. lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,1 + L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] (?,1) 10. start0(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> start(A,C,C,E,E,G,G,I,I,K,K,M,M,A) True (1,1) Signature: {(lbl13,14);(lbl53,14);(lbl91,14);(start,14);(start0,14);(stop,14)} Flow Graph: [0->{},1->{4,5,6},2->{4,5,6},3->{7,8,9},4->{3},5->{},6->{},7->{},8->{4,5,6},9->{4,5,6},10->{0,1,2}] + Applied Processor: FromIts + Details: () * Step 3: Decompose YES + Considered Problem: Rules: start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,0,M,N) [1 >= A && B = C && D = E && F = G && H = I && J = K && L = M && N = A] start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,0,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,1,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] lbl91(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl13(A,L,C,D,E,F,G,H,I,J,K,1 + L,M,N) [D >= H && D >= 1 && 1 + H >= D && A >= 1 + D && N = A && 1 + L = D && J = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl91(A,B,C,D,E,O,G,H,I,J,K,L,M,N) [D >= H && D >= 1 && A >= 1 + D && 1 + H >= D && J = A && 1 + L = D && N = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,H,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,J,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,L,M,N) [2 + H >= A && A >= 2 && A >= 1 + H && 1 + L = A && 2 + B = A && 1 + D = A && N = A && J = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,1 + L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] start0(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> start(A,C,C,E,E,G,G,I,I,K,K,M,M,A) True Signature: {(lbl13,14);(lbl53,14);(lbl91,14);(start,14);(start0,14);(stop,14)} Rule Graph: [0->{},1->{4,5,6},2->{4,5,6},3->{7,8,9},4->{3},5->{},6->{},7->{},8->{4,5,6},9->{4,5,6},10->{0,1,2}] + Applied Processor: Decompose NoGreedy + Details: We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10] | `- p:[4,8,3,9] c: [9] | `- p:[3,4,8] c: [3,4,8] * Step 4: CloseWith YES + Considered Problem: (Rules: start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,0,M,N) [1 >= A && B = C && D = E && F = G && H = I && J = K && L = M && N = A] start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,0,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] start(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1,E,F,G,1,I,2,K,0,M,N) [A >= 2 && B = C && D = E && F = G && H = I && J = K && L = M && N = A] lbl91(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl13(A,L,C,D,E,F,G,H,I,J,K,1 + L,M,N) [D >= H && D >= 1 && 1 + H >= D && A >= 1 + D && N = A && 1 + L = D && J = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl91(A,B,C,D,E,O,G,H,I,J,K,L,M,N) [D >= H && D >= 1 && A >= 1 + D && 1 + H >= D && J = A && 1 + L = D && N = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,H,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] lbl53(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,J,E,F,G,J,I,1 + J,K,L,M,N) [A >= 1 + J && D >= H && D >= 1 && A >= J && J >= 1 + D && 1 + H >= D && 1 + L = D && N = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> stop(A,B,C,D,E,F,G,H,I,J,K,L,M,N) [2 + H >= A && A >= 2 && A >= 1 + H && 1 + L = A && 2 + B = A && 1 + D = A && N = A && J = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] lbl13(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> lbl53(A,B,C,1 + L,E,F,G,1 + L,I,2 + L,K,L,M,N) [A >= 3 + B && A >= 2 + B && H >= B && B >= 0 && 1 + B >= H && L = 1 + B && D = 1 + B && N = A && J = A] start0(A,B,C,D,E,F,G,H,I,J,K,L,M,N) -> start(A,C,C,E,E,G,G,I,I,K,K,M,M,A) True Signature: {(lbl13,14);(lbl53,14);(lbl91,14);(start,14);(start0,14);(stop,14)} Rule Graph: [0->{},1->{4,5,6},2->{4,5,6},3->{7,8,9},4->{3},5->{},6->{},7->{},8->{4,5,6},9->{4,5,6},10->{0,1,2}] ,We construct a looptree: P: [0,1,2,3,4,5,6,7,8,9,10] | `- p:[4,8,3,9] c: [9] | `- p:[3,4,8] c: [3,4,8]) + Applied Processor: CloseWith True + Details: () YES