We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { h(X, Z) -> f(X, s(X), Z) , f(X, Y, g(X, Y)) -> h(0(), g(X, Y)) , g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , f^#(X, Y, g(X, Y)) -> c_2(h^#(0(), g(X, Y))) , g^#(X, s(Y)) -> c_3(g^#(X, Y)) , g^#(0(), Y) -> c_4() } 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: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , f^#(X, Y, g(X, Y)) -> c_2(h^#(0(), g(X, Y))) , g^#(X, s(Y)) -> c_3(g^#(X, Y)) , g^#(0(), Y) -> c_4() } Strict Trs: { h(X, Z) -> f(X, s(X), Z) , f(X, Y, g(X, Y)) -> h(0(), g(X, Y)) , g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , f^#(X, Y, g(X, Y)) -> c_2(h^#(0(), g(X, Y))) , g^#(X, s(Y)) -> c_3(g^#(X, Y)) , g^#(0(), Y) -> c_4() } Strict Trs: { g(X, s(Y)) -> g(X, Y) , g(0(), 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_2) = {1}, Uargs(c_3) = {1} TcT has computed the following constructor-restricted matrix interpretation. [s](x1) = [0 0] x1 + [0] [0 1] [2] [g](x1, x2) = [0 1] x2 + [1] [0 0] [0] [0] = [0] [0] [h^#](x1, x2) = [0] [0] [c_1](x1) = [1 0] x1 + [0] [0 1] [0] [f^#](x1, x2, x3) = [0] [0] [c_2](x1) = [1 0] x1 + [0] [0 1] [0] [g^#](x1, x2) = [0] [0] [c_3](x1) = [1 0] x1 + [0] [0 1] [0] [c_4] = [0] [0] The order satisfies the following ordering constraints: [g(X, s(Y))] = [0 1] Y + [3] [0 0] [0] > [0 1] Y + [1] [0 0] [0] = [g(X, Y)] [g(0(), Y)] = [0 1] Y + [1] [0 0] [0] > [0] [0] = [0()] [h^#(X, Z)] = [0] [0] >= [0] [0] = [c_1(f^#(X, s(X), Z))] [f^#(X, Y, g(X, Y))] = [0] [0] >= [0] [0] = [c_2(h^#(0(), g(X, Y)))] [g^#(X, s(Y))] = [0] [0] >= [0] [0] = [c_3(g^#(X, Y))] [g^#(0(), Y)] = [0] [0] >= [0] [0] = [c_4()] 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: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , f^#(X, Y, g(X, Y)) -> c_2(h^#(0(), g(X, Y))) , g^#(X, s(Y)) -> c_3(g^#(X, Y)) , g^#(0(), Y) -> c_4() } Weak Trs: { g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph: 1: h^#(X, Z) -> c_1(f^#(X, s(X), Z)) 2: f^#(X, Y, g(X, Y)) -> c_2(h^#(0(), g(X, Y))) -->_1 h^#(X, Z) -> c_1(f^#(X, s(X), Z)) :1 3: g^#(X, s(Y)) -> c_3(g^#(X, Y)) -->_1 g^#(0(), Y) -> c_4() :4 -->_1 g^#(X, s(Y)) -> c_3(g^#(X, Y)) :3 4: g^#(0(), Y) -> c_4() Only the nodes {1,3,4} are reachable from nodes {1,3,4} 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: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , g^#(X, s(Y)) -> c_3(g^#(X, Y)) , g^#(0(), Y) -> c_4() } Weak Trs: { g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,3} by applications of Pre({1,3}) = {2}. Here rules are labeled as follows: DPs: { 1: h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , 2: g^#(X, s(Y)) -> c_3(g^#(X, Y)) , 3: g^#(0(), Y) -> c_4() } 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_3(g^#(X, Y)) } Weak DPs: { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , g^#(0(), Y) -> c_4() } Weak Trs: { g(X, s(Y)) -> g(X, Y) , g(0(), Y) -> 0() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { h^#(X, Z) -> c_1(f^#(X, s(X), Z)) , g^#(0(), Y) -> c_4() } 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_3(g^#(X, Y)) } Weak Trs: { g(X, s(Y)) -> g(X, Y) , g(0(), 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_3(g^#(X, Y)) } 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^#(X, s(Y)) -> c_3(g^#(X, Y)) } 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^#) = {1}, safe(c_3) = {} 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, 2], pi(c_3) = [1] Usable defined function symbols are a subset of: {g^#} For your convenience, here are the satisfied ordering constraints: pi(g^#(X, s(Y))) = g^#(s(; Y); X) > c_3(g^#(Y; X);) = pi(c_3(g^#(X, Y))) 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^#(X, s(Y)) -> c_3(g^#(X, Y)) } 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^#(X, s(Y)) -> c_3(g^#(X, Y)) } 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))