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