YES(?,O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { f(x, 0()) -> s(0()) , f(s(x), s(y)) -> s(f(x, y)) , g(0(), x) -> g(f(x, x), x) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We add following dependency tuples: Strict DPs: { f^#(x, 0()) -> c_1() , f^#(s(x), s(y)) -> c_2(f^#(x, y)) , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, 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, 0()) -> c_1() , f^#(s(x), s(y)) -> c_2(f^#(x, y)) , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) } Weak Trs: { f(x, 0()) -> s(0()) , f(s(x), s(y)) -> s(f(x, y)) , g(0(), x) -> g(f(x, x), x) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1} by applications of Pre({1}) = {2,3}. Here rules are labeled as follows: DPs: { 1: f^#(x, 0()) -> c_1() , 2: f^#(s(x), s(y)) -> c_2(f^#(x, y)) , 3: g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(s(x), s(y)) -> c_2(f^#(x, y)) , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) } Weak DPs: { f^#(x, 0()) -> c_1() } Weak Trs: { f(x, 0()) -> s(0()) , f(s(x), s(y)) -> s(f(x, y)) , g(0(), x) -> g(f(x, x), 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, 0()) -> c_1() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(s(x), s(y)) -> c_2(f^#(x, y)) , g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) } Weak Trs: { f(x, 0()) -> s(0()) , f(s(x), s(y)) -> s(f(x, y)) , g(0(), x) -> g(f(x, x), 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: { g^#(0(), x) -> c_3(g^#(f(x, x), x), f^#(x, x)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(s(x), s(y)) -> c_1(f^#(x, y)) , g^#(0(), x) -> c_2(f^#(x, x)) } Weak Trs: { f(x, 0()) -> s(0()) , f(s(x), s(y)) -> s(f(x, y)) , g(0(), x) -> g(f(x, x), 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: { f^#(s(x), s(y)) -> c_1(f^#(x, y)) , g^#(0(), x) -> c_2(f^#(x, x)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Consider the dependency graph 1: f^#(s(x), s(y)) -> c_1(f^#(x, y)) -->_1 f^#(s(x), s(y)) -> c_1(f^#(x, y)) :1 2: g^#(0(), x) -> c_2(f^#(x, x)) -->_1 f^#(s(x), s(y)) -> c_1(f^#(x, y)) :1 Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { g^#(0(), x) -> c_2(f^#(x, x)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(s(x), s(y)) -> c_1(f^#(x, y)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following argument positions are usable: Uargs(c_1) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [s](x1) = [1] x1 + [2] [f^#](x1, x2) = [2] x1 + [0] [c_1](x1) = [1] x1 + [1] This order satisfies following ordering constraints: [f^#(s(x), s(y))] = [2] x + [4] > [2] x + [1] = [c_1(f^#(x, y))] Hurray, we answered YES(?,O(n^1))