MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__f(X) -> cons(mark(X), f(g(X))) , a__f(X) -> f(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(g(X)) -> a__g(mark(X)) , mark(0()) -> 0() , mark(s(X)) -> s(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , a__g(X) -> g(X) , a__g(0()) -> s(0()) , a__g(s(X)) -> s(s(a__g(mark(X)))) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a__f^#(X) -> c_1(mark^#(X)) , a__f^#(X) -> c_2() , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(f(X)) -> c_4(a__f^#(mark(X)), mark^#(X)) , mark^#(g(X)) -> c_5(a__g^#(mark(X)), mark^#(X)) , mark^#(0()) -> c_6() , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(sel(X1, X2)) -> c_8(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__g^#(X) -> c_9() , a__g^#(0()) -> c_10() , a__g^#(s(X)) -> c_11(a__g^#(mark(X)), mark^#(X)) , a__sel^#(X1, X2) -> c_12() , a__sel^#(0(), cons(X, Y)) -> c_13(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__f^#(X) -> c_1(mark^#(X)) , a__f^#(X) -> c_2() , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(f(X)) -> c_4(a__f^#(mark(X)), mark^#(X)) , mark^#(g(X)) -> c_5(a__g^#(mark(X)), mark^#(X)) , mark^#(0()) -> c_6() , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(sel(X1, X2)) -> c_8(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__g^#(X) -> c_9() , a__g^#(0()) -> c_10() , a__g^#(s(X)) -> c_11(a__g^#(mark(X)), mark^#(X)) , a__sel^#(X1, X2) -> c_12() , a__sel^#(0(), cons(X, Y)) -> c_13(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z)) } Weak Trs: { a__f(X) -> cons(mark(X), f(g(X))) , a__f(X) -> f(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(g(X)) -> a__g(mark(X)) , mark(0()) -> 0() , mark(s(X)) -> s(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , a__g(X) -> g(X) , a__g(0()) -> s(0()) , a__g(s(X)) -> s(s(a__g(mark(X)))) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,6,9,10,12} by applications of Pre({2,6,9,10,12}) = {1,3,4,5,7,8,11,13,14}. Here rules are labeled as follows: DPs: { 1: a__f^#(X) -> c_1(mark^#(X)) , 2: a__f^#(X) -> c_2() , 3: mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , 4: mark^#(f(X)) -> c_4(a__f^#(mark(X)), mark^#(X)) , 5: mark^#(g(X)) -> c_5(a__g^#(mark(X)), mark^#(X)) , 6: mark^#(0()) -> c_6() , 7: mark^#(s(X)) -> c_7(mark^#(X)) , 8: mark^#(sel(X1, X2)) -> c_8(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 9: a__g^#(X) -> c_9() , 10: a__g^#(0()) -> c_10() , 11: a__g^#(s(X)) -> c_11(a__g^#(mark(X)), mark^#(X)) , 12: a__sel^#(X1, X2) -> c_12() , 13: a__sel^#(0(), cons(X, Y)) -> c_13(mark^#(X)) , 14: a__sel^#(s(X), cons(Y, Z)) -> c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__f^#(X) -> c_1(mark^#(X)) , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(f(X)) -> c_4(a__f^#(mark(X)), mark^#(X)) , mark^#(g(X)) -> c_5(a__g^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(sel(X1, X2)) -> c_8(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__g^#(s(X)) -> c_11(a__g^#(mark(X)), mark^#(X)) , a__sel^#(0(), cons(X, Y)) -> c_13(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z)) } Weak DPs: { a__f^#(X) -> c_2() , mark^#(0()) -> c_6() , a__g^#(X) -> c_9() , a__g^#(0()) -> c_10() , a__sel^#(X1, X2) -> c_12() } Weak Trs: { a__f(X) -> cons(mark(X), f(g(X))) , a__f(X) -> f(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(g(X)) -> a__g(mark(X)) , mark(0()) -> 0() , mark(s(X)) -> s(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , a__g(X) -> g(X) , a__g(0()) -> s(0()) , a__g(s(X)) -> s(s(a__g(mark(X)))) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) } 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__f^#(X) -> c_2() , mark^#(0()) -> c_6() , a__g^#(X) -> c_9() , a__g^#(0()) -> c_10() , a__sel^#(X1, X2) -> c_12() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__f^#(X) -> c_1(mark^#(X)) , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(f(X)) -> c_4(a__f^#(mark(X)), mark^#(X)) , mark^#(g(X)) -> c_5(a__g^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_7(mark^#(X)) , mark^#(sel(X1, X2)) -> c_8(a__sel^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__g^#(s(X)) -> c_11(a__g^#(mark(X)), mark^#(X)) , a__sel^#(0(), cons(X, Y)) -> c_13(mark^#(X)) , a__sel^#(s(X), cons(Y, Z)) -> c_14(a__sel^#(mark(X), mark(Z)), mark^#(X), mark^#(Z)) } Weak Trs: { a__f(X) -> cons(mark(X), f(g(X))) , a__f(X) -> f(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(g(X)) -> a__g(mark(X)) , mark(0()) -> 0() , mark(s(X)) -> s(mark(X)) , mark(sel(X1, X2)) -> a__sel(mark(X1), mark(X2)) , a__g(X) -> g(X) , a__g(0()) -> s(0()) , a__g(s(X)) -> s(s(a__g(mark(X)))) , a__sel(X1, X2) -> sel(X1, X2) , a__sel(0(), cons(X, Y)) -> mark(X) , a__sel(s(X), cons(Y, Z)) -> a__sel(mark(X), mark(Z)) } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..