MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(nil()) -> nil() , mark(s(X)) -> s(mark(X)) , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(length(X)) -> a__length(mark(X)) , a__length(X) -> length(X) , a__length(cons(N, L)) -> s(a__length(mark(L))) , a__length(nil()) -> 0() , a__take(X1, X2) -> take(X1, X2) , a__take(0(), IL) -> nil() , a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__and^#(X1, X2) -> c_3() , a__and^#(tt(), X) -> c_4(mark^#(X)) , mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , mark^#(0()) -> c_6() , mark^#(zeros()) -> c_7(a__zeros^#()) , mark^#(tt()) -> c_8() , mark^#(nil()) -> c_9() , mark^#(s(X)) -> c_10(mark^#(X)) , mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , a__take^#(X1, X2) -> c_17() , a__take^#(0(), IL) -> c_18() , a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , a__length^#(X) -> c_14() , a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) , a__length^#(nil()) -> c_16() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__and^#(X1, X2) -> c_3() , a__and^#(tt(), X) -> c_4(mark^#(X)) , mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , mark^#(0()) -> c_6() , mark^#(zeros()) -> c_7(a__zeros^#()) , mark^#(tt()) -> c_8() , mark^#(nil()) -> c_9() , mark^#(s(X)) -> c_10(mark^#(X)) , mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , a__take^#(X1, X2) -> c_17() , a__take^#(0(), IL) -> c_18() , a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , a__length^#(X) -> c_14() , a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) , a__length^#(nil()) -> c_16() } Weak Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(nil()) -> nil() , mark(s(X)) -> s(mark(X)) , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(length(X)) -> a__length(mark(X)) , a__length(X) -> length(X) , a__length(cons(N, L)) -> s(a__length(mark(L))) , a__length(nil()) -> 0() , a__take(X1, X2) -> take(X1, X2) , a__take(0(), IL) -> nil() , a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {1,2,3,6,8,9,14,15,17,19} by applications of Pre({1,2,3,6,8,9,14,15,17,19}) = {4,5,7,10,11,12,13,16,18}. Here rules are labeled as follows: DPs: { 1: a__zeros^#() -> c_1() , 2: a__zeros^#() -> c_2() , 3: a__and^#(X1, X2) -> c_3() , 4: a__and^#(tt(), X) -> c_4(mark^#(X)) , 5: mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , 6: mark^#(0()) -> c_6() , 7: mark^#(zeros()) -> c_7(a__zeros^#()) , 8: mark^#(tt()) -> c_8() , 9: mark^#(nil()) -> c_9() , 10: mark^#(s(X)) -> c_10(mark^#(X)) , 11: mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 12: mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , 13: mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , 14: a__take^#(X1, X2) -> c_17() , 15: a__take^#(0(), IL) -> c_18() , 16: a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , 17: a__length^#(X) -> c_14() , 18: a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) , 19: a__length^#(nil()) -> c_16() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__and^#(tt(), X) -> c_4(mark^#(X)) , mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , mark^#(zeros()) -> c_7(a__zeros^#()) , mark^#(s(X)) -> c_10(mark^#(X)) , mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) } Weak DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__and^#(X1, X2) -> c_3() , mark^#(0()) -> c_6() , mark^#(tt()) -> c_8() , mark^#(nil()) -> c_9() , a__take^#(X1, X2) -> c_17() , a__take^#(0(), IL) -> c_18() , a__length^#(X) -> c_14() , a__length^#(nil()) -> c_16() } Weak Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(nil()) -> nil() , mark(s(X)) -> s(mark(X)) , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(length(X)) -> a__length(mark(X)) , a__length(X) -> length(X) , a__length(cons(N, L)) -> s(a__length(mark(L))) , a__length(nil()) -> 0() , a__take(X1, X2) -> take(X1, X2) , a__take(0(), IL) -> nil() , a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {3} by applications of Pre({3}) = {1,2,4,5,6,7,8,9}. Here rules are labeled as follows: DPs: { 1: a__and^#(tt(), X) -> c_4(mark^#(X)) , 2: mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , 3: mark^#(zeros()) -> c_7(a__zeros^#()) , 4: mark^#(s(X)) -> c_10(mark^#(X)) , 5: mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 6: mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , 7: mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , 8: a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , 9: a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) , 10: a__zeros^#() -> c_1() , 11: a__zeros^#() -> c_2() , 12: a__and^#(X1, X2) -> c_3() , 13: mark^#(0()) -> c_6() , 14: mark^#(tt()) -> c_8() , 15: mark^#(nil()) -> c_9() , 16: a__take^#(X1, X2) -> c_17() , 17: a__take^#(0(), IL) -> c_18() , 18: a__length^#(X) -> c_14() , 19: a__length^#(nil()) -> c_16() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__and^#(tt(), X) -> c_4(mark^#(X)) , mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , mark^#(s(X)) -> c_10(mark^#(X)) , mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) } Weak DPs: { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__and^#(X1, X2) -> c_3() , mark^#(0()) -> c_6() , mark^#(zeros()) -> c_7(a__zeros^#()) , mark^#(tt()) -> c_8() , mark^#(nil()) -> c_9() , a__take^#(X1, X2) -> c_17() , a__take^#(0(), IL) -> c_18() , a__length^#(X) -> c_14() , a__length^#(nil()) -> c_16() } Weak Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(nil()) -> nil() , mark(s(X)) -> s(mark(X)) , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(length(X)) -> a__length(mark(X)) , a__length(X) -> length(X) , a__length(cons(N, L)) -> s(a__length(mark(L))) , a__length(nil()) -> 0() , a__take(X1, X2) -> take(X1, X2) , a__take(0(), IL) -> nil() , a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { a__zeros^#() -> c_1() , a__zeros^#() -> c_2() , a__and^#(X1, X2) -> c_3() , mark^#(0()) -> c_6() , mark^#(zeros()) -> c_7(a__zeros^#()) , mark^#(tt()) -> c_8() , mark^#(nil()) -> c_9() , a__take^#(X1, X2) -> c_17() , a__take^#(0(), IL) -> c_18() , a__length^#(X) -> c_14() , a__length^#(nil()) -> c_16() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__and^#(tt(), X) -> c_4(mark^#(X)) , mark^#(cons(X1, X2)) -> c_5(mark^#(X1)) , mark^#(s(X)) -> c_10(mark^#(X)) , mark^#(take(X1, X2)) -> c_11(a__take^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_12(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(length(X)) -> c_13(a__length^#(mark(X)), mark^#(X)) , a__take^#(s(M), cons(N, IL)) -> c_19(mark^#(N)) , a__length^#(cons(N, L)) -> c_15(a__length^#(mark(L)), mark^#(L)) } Weak Trs: { a__zeros() -> cons(0(), zeros()) , a__zeros() -> zeros() , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(zeros()) -> a__zeros() , mark(tt()) -> tt() , mark(nil()) -> nil() , mark(s(X)) -> s(mark(X)) , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(length(X)) -> a__length(mark(X)) , a__length(X) -> length(X) , a__length(cons(N, L)) -> s(a__length(mark(L))) , a__length(nil()) -> 0() , a__take(X1, X2) -> take(X1, X2) , a__take(0(), IL) -> nil() , a__take(s(M), cons(N, IL)) -> cons(mark(N), take(M, IL)) } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..