YES(?,O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { a__f(X) -> g(h(f(X))) , a__f(X) -> f(X) , mark(g(X)) -> g(X) , mark(h(X)) -> h(mark(X)) , mark(f(X)) -> a__f(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We add following dependency tuples: Strict DPs: { a__f^#(X) -> c_1() , a__f^#(X) -> c_2() , mark^#(g(X)) -> c_3() , mark^#(h(X)) -> c_4(mark^#(X)) , mark^#(f(X)) -> c_5(a__f^#(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^1)). Strict DPs: { a__f^#(X) -> c_1() , a__f^#(X) -> c_2() , mark^#(g(X)) -> c_3() , mark^#(h(X)) -> c_4(mark^#(X)) , mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) } Weak Trs: { a__f(X) -> g(h(f(X))) , a__f(X) -> f(X) , mark(g(X)) -> g(X) , mark(h(X)) -> h(mark(X)) , mark(f(X)) -> a__f(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,2,3} by applications of Pre({1,2,3}) = {4,5}. Here rules are labeled as follows: DPs: { 1: a__f^#(X) -> c_1() , 2: a__f^#(X) -> c_2() , 3: mark^#(g(X)) -> c_3() , 4: mark^#(h(X)) -> c_4(mark^#(X)) , 5: mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(h(X)) -> c_4(mark^#(X)) , mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) } Weak DPs: { a__f^#(X) -> c_1() , a__f^#(X) -> c_2() , mark^#(g(X)) -> c_3() } Weak Trs: { a__f(X) -> g(h(f(X))) , a__f(X) -> f(X) , mark(g(X)) -> g(X) , mark(h(X)) -> h(mark(X)) , mark(f(X)) -> a__f(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) 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^#(X) -> c_2() , mark^#(g(X)) -> c_3() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(h(X)) -> c_4(mark^#(X)) , mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) } Weak Trs: { a__f(X) -> g(h(f(X))) , a__f(X) -> f(X) , mark(g(X)) -> g(X) , mark(h(X)) -> h(mark(X)) , mark(f(X)) -> a__f(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { mark^#(f(X)) -> c_5(a__f^#(mark(X)), mark^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(h(X)) -> c_1(mark^#(X)) , mark^#(f(X)) -> c_2(mark^#(X)) } Weak Trs: { a__f(X) -> g(h(f(X))) , a__f(X) -> f(X) , mark(g(X)) -> g(X) , mark(h(X)) -> h(mark(X)) , mark(f(X)) -> a__f(mark(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(h(X)) -> c_1(mark^#(X)) , mark^#(f(X)) -> c_2(mark^#(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [h](x1) = [1] x1 + [2] [f](x1) = [1] x1 + [2] [mark^#](x1) = [2] x1 + [0] [c_1](x1) = [1] x1 + [1] [c_2](x1) = [1] x1 + [1] This order satisfies following ordering constraints: [mark^#(h(X))] = [2] X + [4] > [2] X + [1] = [c_1(mark^#(X))] [mark^#(f(X))] = [2] X + [4] > [2] X + [1] = [c_2(mark^#(X))] Hurray, we answered YES(?,O(n^1))