YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { g(x, s(y)) -> g(f(x, y), 0()) , g(0(), f(x, x)) -> x , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0())) , g(s(x), y) -> g(f(x, y), 0()) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) , g^#(0(), f(x, x)) -> c_2() , g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) , g^#(0(), f(x, x)) -> c_2() , g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } Strict Trs: { g(x, s(y)) -> g(f(x, y), 0()) , g(0(), f(x, x)) -> x , g(f(x, y), 0()) -> f(g(x, 0()), g(y, 0())) , g(s(x), y) -> g(f(x, y), 0()) } Obligation: innermost runtime complexity Answer: YES(O(1),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(1),O(n^1)). Strict DPs: { g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) , g^#(0(), f(x, x)) -> c_2() , g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_3) = {1, 2}, Uargs(c_4) = {1} TcT has computed following constructor-restricted matrix interpretation. [0] = [1] [f](x1, x2) = [1] x1 + [1] x2 + [0] [s](x1) = [1] x1 + [0] [g^#](x1, x2) = [1] x1 + [1] x2 + [0] [c_1](x1) = [1] x1 + [0] [c_2] = [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [2] [c_4](x1) = [1] x1 + [0] This order satisfies following ordering constraints: Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) , g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } Weak DPs: { g^#(0(), f(x, x)) -> c_2() } 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. { g^#(0(), f(x, x)) -> c_2() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) , g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Consider the dependency graph 1: g^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) -->_1 g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) :2 2: g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) -->_2 g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) :3 -->_1 g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) :3 -->_2 g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) :2 -->_1 g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) :2 3: g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) -->_1 g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) :2 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^#(x, s(y)) -> c_1(g^#(f(x, y), 0())) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { g^#(f(x, y), 0()) -> c_3(g^#(x, 0()), g^#(y, 0())) , g^#(s(x), y) -> c_4(g^#(f(x, y), 0())) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The following argument positions are usable: Uargs(c_3) = {1, 2}, Uargs(c_4) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [0] = [0] [f](x1, x2) = [1] x1 + [1] x2 + [1] [s](x1) = [1] x1 + [2] [g^#](x1, x2) = [2] x1 + [2] x2 + [0] [c_3](x1, x2) = [1] x1 + [1] x2 + [1] [c_4](x1) = [1] x1 + [1] This order satisfies following ordering constraints: [g^#(f(x, y), 0())] = [2] x + [2] y + [2] > [2] x + [2] y + [1] = [c_3(g^#(x, 0()), g^#(y, 0()))] [g^#(s(x), y)] = [2] x + [2] y + [4] > [2] x + [2] y + [3] = [c_4(g^#(f(x, y), 0()))] Hurray, we answered YES(O(1),O(n^1))