MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__pairNs() -> cons(0(), incr(oddNs())) , a__pairNs() -> pairNs() , a__oddNs() -> oddNs() , a__oddNs() -> a__incr(a__pairNs()) , a__incr(X) -> incr(X) , a__incr(cons(X, XS)) -> cons(s(mark(X)), incr(XS)) , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(0()) -> 0() , mark(incr(X)) -> a__incr(mark(X)) , mark(oddNs()) -> a__oddNs() , mark(s(X)) -> s(mark(X)) , mark(nil()) -> nil() , mark(take(X1, X2)) -> a__take(mark(X1), mark(X2)) , mark(pair(X1, X2)) -> pair(mark(X1), mark(X2)) , mark(zip(X1, X2)) -> a__zip(mark(X1), mark(X2)) , mark(repItems(X)) -> a__repItems(mark(X)) , mark(pairNs()) -> a__pairNs() , mark(tail(X)) -> a__tail(mark(X)) , a__take(X1, X2) -> take(X1, X2) , a__take(0(), XS) -> nil() , a__take(s(N), cons(X, XS)) -> cons(mark(X), take(N, XS)) , a__zip(X1, X2) -> zip(X1, X2) , a__zip(X, nil()) -> nil() , a__zip(cons(X, XS), cons(Y, YS)) -> cons(pair(mark(X), mark(Y)), zip(XS, YS)) , a__zip(nil(), XS) -> nil() , a__tail(X) -> tail(X) , a__tail(cons(X, XS)) -> mark(XS) , a__repItems(X) -> repItems(X) , a__repItems(cons(X, XS)) -> cons(mark(X), cons(X, repItems(XS))) , a__repItems(nil()) -> nil() } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..