MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { a____(X1, X2) -> __(X1, X2) , a____(X, nil()) -> mark(X) , a____(__(X, Y), Z) -> a____(mark(X), a____(mark(Y), mark(Z))) , a____(nil(), X) -> mark(X) , mark(__(X1, X2)) -> a____(mark(X1), mark(X2)) , mark(nil()) -> nil() , mark(tt()) -> tt() , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNePal(X)) -> a__isNePal(mark(X)) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNePal(X) -> isNePal(X) , a__isNePal(__(I, __(P, I))) -> tt() } Obligation: innermost runtime complexity Answer: MAYBE We add following dependency tuples: Strict DPs: { a____^#(X1, X2) -> c_1() , a____^#(X, nil()) -> c_2(mark^#(X)) , a____^#(__(X, Y), Z) -> c_3(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , a____^#(nil(), X) -> c_4(mark^#(X)) , mark^#(__(X1, X2)) -> c_5(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(nil()) -> c_6() , mark^#(tt()) -> c_7() , mark^#(and(X1, X2)) -> c_8(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) , a__and^#(X1, X2) -> c_10() , a__and^#(tt(), X) -> c_11(mark^#(X)) , a__isNePal^#(X) -> c_12() , a__isNePal^#(__(I, __(P, I))) -> c_13() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a____^#(X1, X2) -> c_1() , a____^#(X, nil()) -> c_2(mark^#(X)) , a____^#(__(X, Y), Z) -> c_3(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , a____^#(nil(), X) -> c_4(mark^#(X)) , mark^#(__(X1, X2)) -> c_5(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(nil()) -> c_6() , mark^#(tt()) -> c_7() , mark^#(and(X1, X2)) -> c_8(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) , a__and^#(X1, X2) -> c_10() , a__and^#(tt(), X) -> c_11(mark^#(X)) , a__isNePal^#(X) -> c_12() , a__isNePal^#(__(I, __(P, I))) -> c_13() } Weak Trs: { a____(X1, X2) -> __(X1, X2) , a____(X, nil()) -> mark(X) , a____(__(X, Y), Z) -> a____(mark(X), a____(mark(Y), mark(Z))) , a____(nil(), X) -> mark(X) , mark(__(X1, X2)) -> a____(mark(X1), mark(X2)) , mark(nil()) -> nil() , mark(tt()) -> tt() , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNePal(X)) -> a__isNePal(mark(X)) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNePal(X) -> isNePal(X) , a__isNePal(__(I, __(P, I))) -> tt() } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {1,6,7,10,12,13} by applications of Pre({1,6,7,10,12,13}) = {2,3,4,5,8,9,11}. Here rules are labeled as follows: DPs: { 1: a____^#(X1, X2) -> c_1() , 2: a____^#(X, nil()) -> c_2(mark^#(X)) , 3: a____^#(__(X, Y), Z) -> c_3(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , 4: a____^#(nil(), X) -> c_4(mark^#(X)) , 5: mark^#(__(X1, X2)) -> c_5(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , 6: mark^#(nil()) -> c_6() , 7: mark^#(tt()) -> c_7() , 8: mark^#(and(X1, X2)) -> c_8(a__and^#(mark(X1), X2), mark^#(X1)) , 9: mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) , 10: a__and^#(X1, X2) -> c_10() , 11: a__and^#(tt(), X) -> c_11(mark^#(X)) , 12: a__isNePal^#(X) -> c_12() , 13: a__isNePal^#(__(I, __(P, I))) -> c_13() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a____^#(X, nil()) -> c_2(mark^#(X)) , a____^#(__(X, Y), Z) -> c_3(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , a____^#(nil(), X) -> c_4(mark^#(X)) , mark^#(__(X1, X2)) -> c_5(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_8(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) , a__and^#(tt(), X) -> c_11(mark^#(X)) } Weak DPs: { a____^#(X1, X2) -> c_1() , mark^#(nil()) -> c_6() , mark^#(tt()) -> c_7() , a__and^#(X1, X2) -> c_10() , a__isNePal^#(X) -> c_12() , a__isNePal^#(__(I, __(P, I))) -> c_13() } Weak Trs: { a____(X1, X2) -> __(X1, X2) , a____(X, nil()) -> mark(X) , a____(__(X, Y), Z) -> a____(mark(X), a____(mark(Y), mark(Z))) , a____(nil(), X) -> mark(X) , mark(__(X1, X2)) -> a____(mark(X1), mark(X2)) , mark(nil()) -> nil() , mark(tt()) -> tt() , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNePal(X)) -> a__isNePal(mark(X)) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNePal(X) -> isNePal(X) , a__isNePal(__(I, __(P, I))) -> tt() } 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____^#(X1, X2) -> c_1() , mark^#(nil()) -> c_6() , mark^#(tt()) -> c_7() , a__and^#(X1, X2) -> c_10() , a__isNePal^#(X) -> c_12() , a__isNePal^#(__(I, __(P, I))) -> c_13() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a____^#(X, nil()) -> c_2(mark^#(X)) , a____^#(__(X, Y), Z) -> c_3(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , a____^#(nil(), X) -> c_4(mark^#(X)) , mark^#(__(X1, X2)) -> c_5(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_8(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) , a__and^#(tt(), X) -> c_11(mark^#(X)) } Weak Trs: { a____(X1, X2) -> __(X1, X2) , a____(X, nil()) -> mark(X) , a____(__(X, Y), Z) -> a____(mark(X), a____(mark(Y), mark(Z))) , a____(nil(), X) -> mark(X) , mark(__(X1, X2)) -> a____(mark(X1), mark(X2)) , mark(nil()) -> nil() , mark(tt()) -> tt() , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNePal(X)) -> a__isNePal(mark(X)) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNePal(X) -> isNePal(X) , a__isNePal(__(I, __(P, I))) -> tt() } Obligation: innermost runtime complexity Answer: MAYBE Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { mark^#(isNePal(X)) -> c_9(a__isNePal^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { a____^#(X, nil()) -> c_1(mark^#(X)) , a____^#(__(X, Y), Z) -> c_2(a____^#(mark(X), a____(mark(Y), mark(Z))), mark^#(X), a____^#(mark(Y), mark(Z)), mark^#(Y), mark^#(Z)) , a____^#(nil(), X) -> c_3(mark^#(X)) , mark^#(__(X1, X2)) -> c_4(a____^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2)) , mark^#(and(X1, X2)) -> c_5(a__and^#(mark(X1), X2), mark^#(X1)) , mark^#(isNePal(X)) -> c_6(mark^#(X)) , a__and^#(tt(), X) -> c_7(mark^#(X)) } Weak Trs: { a____(X1, X2) -> __(X1, X2) , a____(X, nil()) -> mark(X) , a____(__(X, Y), Z) -> a____(mark(X), a____(mark(Y), mark(Z))) , a____(nil(), X) -> mark(X) , mark(__(X1, X2)) -> a____(mark(X1), mark(X2)) , mark(nil()) -> nil() , mark(tt()) -> tt() , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNePal(X)) -> a__isNePal(mark(X)) , a__and(X1, X2) -> and(X1, X2) , a__and(tt(), X) -> mark(X) , a__isNePal(X) -> isNePal(X) , a__isNePal(__(I, __(P, I))) -> tt() } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'matrices' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'matrix interpretation of dimension 4' failed due to the following reason: Following exception was raised: stack overflow 2) 'matrix interpretation of dimension 3' failed due to the following reason: The input cannot be shown compatible 3) 'matrix interpretation of dimension 3' failed due to the following reason: The input cannot be shown compatible 4) 'matrix interpretation of dimension 2' failed due to the following reason: The input cannot be shown compatible 5) 'matrix interpretation of dimension 2' failed due to the following reason: The input cannot be shown compatible 6) 'matrix interpretation of dimension 1' failed due to the following reason: The input cannot be shown compatible 2) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. Arrrr..