MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(after(X1, X2)) -> a__after(mark(X1), mark(X2)) , a__after(X1, X2) -> after(X1, X2) , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS)) , a__after(0(), XS) -> mark(XS) } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a__from^#(X) -> c_1(mark^#(X)) , a__from^#(X) -> c_2() , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_5(mark^#(X)) , mark^#(0()) -> c_6() , mark^#(after(X1, X2)) -> c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__after^#(X1, X2) -> c_8() , a__after^#(s(N), cons(X, XS)) -> c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS)) , a__after^#(0(), XS) -> c_10(mark^#(XS)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__from^#(X) -> c_1(mark^#(X)) , a__from^#(X) -> c_2() , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_5(mark^#(X)) , mark^#(0()) -> c_6() , mark^#(after(X1, X2)) -> c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__after^#(X1, X2) -> c_8() , a__after^#(s(N), cons(X, XS)) -> c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS)) , a__after^#(0(), XS) -> c_10(mark^#(XS)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(after(X1, X2)) -> a__after(mark(X1), mark(X2)) , a__after(X1, X2) -> after(X1, X2) , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS)) , a__after(0(), XS) -> mark(XS) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,6,8} by applications of Pre({2,6,8}) = {1,3,4,5,7,9,10}. Here rules are labeled as follows: DPs: { 1: a__from^#(X) -> c_1(mark^#(X)) , 2: a__from^#(X) -> c_2() , 3: mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , 4: mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X)) , 5: mark^#(s(X)) -> c_5(mark^#(X)) , 6: mark^#(0()) -> c_6() , 7: mark^#(after(X1, X2)) -> c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 8: a__after^#(X1, X2) -> c_8() , 9: a__after^#(s(N), cons(X, XS)) -> c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS)) , 10: a__after^#(0(), XS) -> c_10(mark^#(XS)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__from^#(X) -> c_1(mark^#(X)) , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_5(mark^#(X)) , mark^#(after(X1, X2)) -> c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__after^#(s(N), cons(X, XS)) -> c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS)) , a__after^#(0(), XS) -> c_10(mark^#(XS)) } Weak DPs: { a__from^#(X) -> c_2() , mark^#(0()) -> c_6() , a__after^#(X1, X2) -> c_8() } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(after(X1, X2)) -> a__after(mark(X1), mark(X2)) , a__after(X1, X2) -> after(X1, X2) , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS)) , a__after(0(), XS) -> mark(XS) } 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__from^#(X) -> c_2() , mark^#(0()) -> c_6() , a__after^#(X1, X2) -> c_8() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a__from^#(X) -> c_1(mark^#(X)) , mark^#(cons(X1, X2)) -> c_3(mark^#(X1)) , mark^#(from(X)) -> c_4(a__from^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_5(mark^#(X)) , mark^#(after(X1, X2)) -> c_7(a__after^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , a__after^#(s(N), cons(X, XS)) -> c_9(a__after^#(mark(N), mark(XS)), mark^#(N), mark^#(XS)) , a__after^#(0(), XS) -> c_10(mark^#(XS)) } Weak Trs: { a__from(X) -> cons(mark(X), from(s(X))) , a__from(X) -> from(X) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(from(X)) -> a__from(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(after(X1, X2)) -> a__after(mark(X1), mark(X2)) , a__after(X1, X2) -> after(X1, X2) , a__after(s(N), cons(X, XS)) -> a__after(mark(N), mark(XS)) , a__after(0(), XS) -> mark(XS) } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..