We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { f(s(0()), g(x)) -> f(x, g(x)) , g(s(x)) -> g(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { f^#(s(0()), g(x)) -> c_1(f^#(x, g(x))) , g^#(s(x)) -> c_2(g^#(x)) } 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: { f^#(s(0()), g(x)) -> c_1(f^#(x, g(x))) , g^#(s(x)) -> c_2(g^#(x)) } Strict Trs: { f(s(0()), g(x)) -> f(x, g(x)) , g(s(x)) -> g(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { g(s(x)) -> g(x) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { f^#(s(0()), g(x)) -> c_1(f^#(x, g(x))) , g^#(s(x)) -> c_2(g^#(x)) } Strict Trs: { g(s(x)) -> g(x) } 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_2) = {1} TcT has computed the following constructor-restricted matrix interpretation. [s](x1) = [0 0] x1 + [0] [0 1] [2] [0] = [0] [0] [g](x1) = [0 1] x1 + [0] [0 0] [0] [f^#](x1, x2) = [0] [0] [c_1](x1) = [1 0] x1 + [0] [0 1] [0] [g^#](x1) = [0] [0] [c_2](x1) = [1 0] x1 + [0] [0 1] [0] The order satisfies the following ordering constraints: [g(s(x))] = [0 1] x + [2] [0 0] [0] > [0 1] x + [0] [0 0] [0] = [g(x)] [f^#(s(0()), g(x))] = [0] [0] >= [0] [0] = [c_1(f^#(x, g(x)))] [g^#(s(x))] = [0] [0] >= [0] [0] = [c_2(g^#(x))] 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),O(n^1)). Strict DPs: { f^#(s(0()), g(x)) -> c_1(f^#(x, g(x))) , g^#(s(x)) -> c_2(g^#(x)) } Weak Trs: { g(s(x)) -> g(x) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph: 1: f^#(s(0()), g(x)) -> c_1(f^#(x, g(x))) 2: g^#(s(x)) -> c_2(g^#(x)) -->_1 g^#(s(x)) -> c_2(g^#(x)) :2 Only the nodes {2} are reachable from nodes {2} that start derivation from marked basic terms. The nodes not reachable are removed from the problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { g^#(s(x)) -> c_2(g^#(x)) } Weak Trs: { g(s(x)) -> g(x) } 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^#(s(x)) -> c_2(g^#(x)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: g^#(s(x)) -> c_2(g^#(x)) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(s) = {1}, safe(g^#) = {}, safe(c_2) = {} and precedence empty . Following symbols are considered recursive: {g^#} The recursion depth is 1. Further, following argument filtering is employed: pi(s) = [1], pi(g^#) = [1], pi(c_2) = [1] Usable defined function symbols are a subset of: {g^#} For your convenience, here are the satisfied ordering constraints: pi(g^#(s(x))) = g^#(s(; x);) > c_2(g^#(x;);) = pi(c_2(g^#(x))) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { g^#(s(x)) -> c_2(g^#(x)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { g^#(s(x)) -> c_2(g^#(x)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))