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: { f(c(s(x), y)) -> f(c(x, s(y))) , g(c(x, s(y))) -> g(c(s(x), y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { f^#(c(s(x), y)) -> c_1(f^#(c(x, s(y)))) , g^#(c(x, s(y))) -> c_2(g^#(c(s(x), y))) } 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^#(c(s(x), y)) -> c_1(f^#(c(x, s(y)))) , g^#(c(x, s(y))) -> c_2(g^#(c(s(x), y))) } Strict Trs: { f(c(s(x), y)) -> f(c(x, s(y))) , g(c(x, s(y))) -> g(c(s(x), y)) } 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: { f^#(c(s(x), y)) -> c_1(f^#(c(x, s(y)))) , g^#(c(x, s(y))) -> c_2(g^#(c(s(x), y))) } 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 following constructor-restricted matrix interpretation. [c](x1, x2) = [1] x2 + [0] [s](x1) = [1] x1 + [1] [f^#](x1) = [0] [c_1](x1) = [1] x1 + [1] [g^#](x1) = [1] x1 + [0] [c_2](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: { f^#(c(s(x), y)) -> c_1(f^#(c(x, s(y)))) } Weak DPs: { g^#(c(x, s(y))) -> c_2(g^#(c(s(x), y))) } 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^#(c(x, s(y))) -> c_2(g^#(c(s(x), y))) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(c(s(x), y)) -> c_1(f^#(c(x, s(y)))) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS)' as induced by the safe mapping safe(c) = {1, 2}, safe(s) = {1}, safe(f^#) = {}, safe(c_1) = {} and precedence empty . Following symbols are considered recursive: {f^#} The recursion depth is 1. Further, following argument filtering is employed: pi(c) = [1], pi(s) = [1], pi(f^#) = [1], pi(c_1) = [1] Usable defined function symbols are a subset of: {f^#} For your convenience, here are the satisfied ordering constraints: pi(f^#(c(s(x), y))) = f^#(c(; s(; x));) > c_1(f^#(c(; x););) = pi(c_1(f^#(c(x, s(y))))) Hurray, we answered YES(O(1),O(n^1))