MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a__U11(X1, X2, X3) -> U11(X1, X2, X3) , a__U11(tt(), M, N) -> a__U12(tt(), M, N) , a__U12(X1, X2, X3) -> U12(X1, X2, X3) , a__U12(tt(), M, N) -> s(a__plus(mark(N), mark(M))) , a__plus(X1, X2) -> plus(X1, X2) , a__plus(N, s(M)) -> a__U11(tt(), M, N) , a__plus(N, 0()) -> mark(N) , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , mark(U11(X1, X2, X3)) -> a__U11(mark(X1), X2, X3) , mark(U12(X1, X2, X3)) -> a__U12(mark(X1), X2, X3) , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) , mark(U21(X1, X2, X3)) -> a__U21(mark(X1), X2, X3) , mark(U22(X1, X2, X3)) -> a__U22(mark(X1), X2, X3) , mark(x(X1, X2)) -> a__x(mark(X1), mark(X2)) , a__U21(X1, X2, X3) -> U21(X1, X2, X3) , a__U21(tt(), M, N) -> a__U22(tt(), M, N) , a__U22(X1, X2, X3) -> U22(X1, X2, X3) , a__U22(tt(), M, N) -> a__plus(a__x(mark(N), mark(M)), mark(N)) , a__x(X1, X2) -> x(X1, X2) , a__x(N, s(M)) -> a__U21(tt(), M, N) , a__x(N, 0()) -> 0() } Obligation: innermost runtime complexity Answer: MAYBE The input cannot be shown compatible Arrrr..