YES(?,O(n^2)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^2)). Strict Trs: { a__f(X) -> f(X) , a__f(0()) -> cons(0(), f(s(0()))) , a__f(s(0())) -> a__f(a__p(s(0()))) , a__p(X) -> p(X) , a__p(s(0())) -> 0() , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(p(X)) -> a__p(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^2)) We add following dependency tuples: Strict DPs: { a__f^#(X) -> c_1() , a__f^#(0()) -> c_2() , a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , a__p^#(X) -> c_4() , a__p^#(s(0())) -> c_5() , mark^#(0()) -> c_6() , mark^#(cons(X1, X2)) -> c_7(mark^#(X1)) , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(?,O(n^2)). Strict DPs: { a__f^#(X) -> c_1() , a__f^#(0()) -> c_2() , a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , a__p^#(X) -> c_4() , a__p^#(s(0())) -> c_5() , mark^#(0()) -> c_6() , mark^#(cons(X1, X2)) -> c_7(mark^#(X1)) , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } Weak Trs: { a__f(X) -> f(X) , a__f(0()) -> cons(0(), f(s(0()))) , a__f(s(0())) -> a__f(a__p(s(0()))) , a__p(X) -> p(X) , a__p(s(0())) -> 0() , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(p(X)) -> a__p(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^2)) We estimate the number of application of {1,2,4,5,6} by applications of Pre({1,2,4,5,6}) = {3,7,8,9,10}. Here rules are labeled as follows: DPs: { 1: a__f^#(X) -> c_1() , 2: a__f^#(0()) -> c_2() , 3: a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , 4: a__p^#(X) -> c_4() , 5: a__p^#(s(0())) -> c_5() , 6: mark^#(0()) -> c_6() , 7: mark^#(cons(X1, X2)) -> c_7(mark^#(X1)) , 8: mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X)) , 9: mark^#(s(X)) -> c_9(mark^#(X)) , 10: mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^2)). Strict DPs: { a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1)) , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } Weak DPs: { a__f^#(X) -> c_1() , a__f^#(0()) -> c_2() , a__p^#(X) -> c_4() , a__p^#(s(0())) -> c_5() , mark^#(0()) -> c_6() } Weak Trs: { a__f(X) -> f(X) , a__f(0()) -> cons(0(), f(s(0()))) , a__f(s(0())) -> a__f(a__p(s(0()))) , a__p(X) -> p(X) , a__p(s(0())) -> 0() , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(p(X)) -> a__p(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^2)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { a__f^#(X) -> c_1() , a__f^#(0()) -> c_2() , a__p^#(X) -> c_4() , a__p^#(s(0())) -> c_5() , mark^#(0()) -> c_6() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^2)). Strict DPs: { a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , mark^#(cons(X1, X2)) -> c_7(mark^#(X1)) , mark^#(f(X)) -> c_8(a__f^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_9(mark^#(X)) , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } Weak Trs: { a__f(X) -> f(X) , a__f(0()) -> cons(0(), f(s(0()))) , a__f(s(0())) -> a__f(a__p(s(0()))) , a__p(X) -> p(X) , a__p(s(0())) -> 0() , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(p(X)) -> a__p(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^2)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { a__f^#(s(0())) -> c_3(a__f^#(a__p(s(0()))), a__p^#(s(0()))) , mark^#(p(X)) -> c_10(a__p^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^2)). Strict DPs: { a__f^#(s(0())) -> c_1(a__f^#(a__p(s(0())))) , mark^#(cons(X1, X2)) -> c_2(mark^#(X1)) , mark^#(f(X)) -> c_3(a__f^#(mark(X)), mark^#(X)) , mark^#(s(X)) -> c_4(mark^#(X)) , mark^#(p(X)) -> c_5(mark^#(X)) } Weak Trs: { a__f(X) -> f(X) , a__f(0()) -> cons(0(), f(s(0()))) , a__f(s(0())) -> a__f(a__p(s(0()))) , a__p(X) -> p(X) , a__p(s(0())) -> 0() , mark(0()) -> 0() , mark(cons(X1, X2)) -> cons(mark(X1), X2) , mark(f(X)) -> a__f(mark(X)) , mark(s(X)) -> s(mark(X)) , mark(p(X)) -> a__p(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^2)) The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1, 2}, Uargs(c_4) = {1}, Uargs(c_5) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [a__f](x1) = [1 3] x1 + [0] [0 0] [2] [0] = [2] [0] [cons](x1, x2) = [1 2] x1 + [0] [0 0] [1] [f](x1) = [1 2] x1 + [0] [0 0] [2] [s](x1) = [1 2] x1 + [0] [0 0] [1] [a__p](x1) = [1 2] x1 + [1] [0 0] [0] [mark](x1) = [2 0] x1 + [0] [0 1] [0] [p](x1) = [1 2] x1 + [1] [0 0] [0] [a__f^#](x1) = [0 1] x1 + [0] [0 0] [0] [mark^#](x1) = [2 2] x1 + [0] [0 1] [0] [c_1](x1) = [2 0] x1 + [0] [0 0] [0] [c_2](x1) = [1 0] x1 + [1] [0 0] [1] [c_3](x1, x2) = [1 0] x1 + [1 1] x2 + [3] [0 0] [0 0] [1] [c_4](x1) = [1 0] x1 + [1] [0 0] [1] [c_5](x1) = [1 0] x1 + [1] [0 0] [0] This order satisfies following ordering constraints: [a__f(X)] = [1 3] X + [0] [0 0] [2] >= [1 2] X + [0] [0 0] [2] = [f(X)] [a__f(0())] = [2] [2] >= [2] [1] = [cons(0(), f(s(0())))] [a__f(s(0()))] = [5] [2] >= [5] [2] = [a__f(a__p(s(0())))] [a__p(X)] = [1 2] X + [1] [0 0] [0] >= [1 2] X + [1] [0 0] [0] = [p(X)] [a__p(s(0()))] = [5] [0] > [2] [0] = [0()] [mark(0())] = [4] [0] > [2] [0] = [0()] [mark(cons(X1, X2))] = [2 4] X1 + [0] [0 0] [1] >= [2 2] X1 + [0] [0 0] [1] = [cons(mark(X1), X2)] [mark(f(X))] = [2 4] X + [0] [0 0] [2] >= [2 3] X + [0] [0 0] [2] = [a__f(mark(X))] [mark(s(X))] = [2 4] X + [0] [0 0] [1] >= [2 2] X + [0] [0 0] [1] = [s(mark(X))] [mark(p(X))] = [2 4] X + [2] [0 0] [0] > [2 2] X + [1] [0 0] [0] = [a__p(mark(X))] [a__f^#(s(0()))] = [1] [0] > [0] [0] = [c_1(a__f^#(a__p(s(0()))))] [mark^#(cons(X1, X2))] = [2 4] X1 + [2] [0 0] [1] > [2 2] X1 + [1] [0 0] [1] = [c_2(mark^#(X1))] [mark^#(f(X))] = [2 4] X + [4] [0 0] [2] > [2 4] X + [3] [0 0] [1] = [c_3(a__f^#(mark(X)), mark^#(X))] [mark^#(s(X))] = [2 4] X + [2] [0 0] [1] > [2 2] X + [1] [0 0] [1] = [c_4(mark^#(X))] [mark^#(p(X))] = [2 4] X + [2] [0 0] [0] > [2 2] X + [1] [0 0] [0] = [c_5(mark^#(X))] Hurray, we answered YES(?,O(n^2))