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(0()) -> 0() , g(s(x)) -> f(g(x)) , f(0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { g^#(0()) -> c_1() , g^#(s(x)) -> c_2(f^#(g(x))) , f^#(0()) -> c_3() } 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^#(0()) -> c_1() , g^#(s(x)) -> c_2(f^#(g(x))) , f^#(0()) -> c_3() } Strict Trs: { g(0()) -> 0() , g(s(x)) -> f(g(x)) , f(0()) -> 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(f) = {1}, Uargs(c_2) = {1}, Uargs(f^#) = {1} TcT has computed following constructor-restricted matrix interpretation. [g](x1) = [2] x1 + [0] [0] = [2] [s](x1) = [1] x1 + [2] [f](x1) = [1] x1 + [1] [g^#](x1) = [2] x1 + [2] [c_1] = [1] [c_2](x1) = [1] x1 + [2] [f^#](x1) = [1] x1 + [2] [c_3] = [1] This order satisfies following ordering constraints: [g(0())] = [4] > [2] = [0()] [g(s(x))] = [2] x + [4] > [2] x + [1] = [f(g(x))] [f(0())] = [3] > [2] = [0()] 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(1)). Weak DPs: { g^#(0()) -> c_1() , g^#(s(x)) -> c_2(f^#(g(x))) , f^#(0()) -> c_3() } Weak Trs: { g(0()) -> 0() , g(s(x)) -> f(g(x)) , f(0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(?,O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { g^#(0()) -> c_1() , g^#(s(x)) -> c_2(f^#(g(x))) , f^#(0()) -> c_3() } We are left with following problem, upon which TcT provides the certificate YES(?,O(1)). Weak Trs: { g(0()) -> 0() , g(s(x)) -> f(g(x)) , f(0()) -> 0() } Obligation: innermost runtime complexity Answer: YES(?,O(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)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(?,O(1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS)' as induced by the safe mapping and precedence empty . Following symbols are considered recursive: {} The recursion depth is 0. Further, following argument filtering is employed: empty Usable defined function symbols are a subset of: {} For your convenience, here are the satisfied ordering constraints: Hurray, we answered YES(O(1),O(n^1))