YES(?,O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We add following dependency tuples: Strict DPs: { f^#(X) -> c_1() , f^#(X) -> c_2() , h^#(X) -> c_3() , activate^#(X) -> c_4() , activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(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: { f^#(X) -> c_1() , f^#(X) -> c_2() , h^#(X) -> c_3() , activate^#(X) -> c_4() , activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) } Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,2,3,4} by applications of Pre({1,2,3,4}) = {5,6}. Here rules are labeled as follows: DPs: { 1: f^#(X) -> c_1() , 2: f^#(X) -> c_2() , 3: h^#(X) -> c_3() , 4: activate^#(X) -> c_4() , 5: activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , 6: activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) } Weak DPs: { f^#(X) -> c_1() , f^#(X) -> c_2() , h^#(X) -> c_3() , activate^#(X) -> c_4() } Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(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. { f^#(X) -> c_1() , f^#(X) -> c_2() , h^#(X) -> c_3() , activate^#(X) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) } Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(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: { activate^#(n__h(X)) -> c_5(h^#(activate(X)), activate^#(X)) , activate^#(n__f(X)) -> c_6(f^#(activate(X)), activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { activate^#(n__h(X)) -> c_1(activate^#(X)) , activate^#(n__f(X)) -> c_2(activate^#(X)) } Weak Trs: { f(X) -> g(n__h(n__f(X))) , f(X) -> n__f(X) , h(X) -> n__h(X) , activate(X) -> X , activate(n__h(X)) -> h(activate(X)) , activate(n__f(X)) -> f(activate(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: { activate^#(n__h(X)) -> c_1(activate^#(X)) , activate^#(n__f(X)) -> c_2(activate^#(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). [n__h](x1) = [1] x1 + [2] [n__f](x1) = [1] x1 + [2] [activate^#](x1) = [2] x1 + [0] [c_1](x1) = [1] x1 + [1] [c_2](x1) = [1] x1 + [1] This order satisfies following ordering constraints: [activate^#(n__h(X))] = [2] X + [4] > [2] X + [1] = [c_1(activate^#(X))] [activate^#(n__f(X))] = [2] X + [4] > [2] X + [1] = [c_2(activate^#(X))] Hurray, we answered YES(?,O(n^1))