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: { perfectp(0()) -> false() , perfectp(s(x)) -> f(x, s(0()), s(x), s(x)) , f(0(), y, 0(), u) -> true() , f(0(), y, s(z), u) -> false() , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u) , f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } 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: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Strict Trs: { perfectp(0()) -> false() , perfectp(s(x)) -> f(x, s(0()), s(x), s(x)) , f(0(), y, 0(), u) -> true() , f(0(), y, s(z), u) -> false() , f(s(x), 0(), z, u) -> f(x, u, minus(z, s(x)), u) , f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) } 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: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } 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_2) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {2} TcT has computed following constructor-restricted matrix interpretation. [0] = [0] [s](x1) = [0] [minus](x1, x2) = [1] x1 + [2] [perfectp^#](x1) = [0] [c_1] = [1] [c_2](x1) = [1] x1 + [0] [f^#](x1, x2, x3, x4) = [1] [c_3] = [0] [c_4] = [0] [c_5](x1) = [1] x1 + [0] [c_6](x1, x2) = [1] x2 + [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: { perfectp^#(0()) -> c_1() , perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Weak DPs: { f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1} by applications of Pre({1}) = {}. Here rules are labeled as follows: DPs: { 1: perfectp^#(0()) -> c_1() , 2: perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , 3: f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , 4: f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) , 5: f^#(0(), y, 0(), u) -> c_3() , 6: f^#(0(), y, s(z), u) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Weak DPs: { perfectp^#(0()) -> c_1() , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } 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. { perfectp^#(0()) -> c_1() , f^#(0(), y, 0(), u) -> c_3() , f^#(0(), y, s(z), u) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_2(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_5(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { f^#(s(x), s(y), z, u) -> c_6(f^#(s(x), minus(y, x), z, u), f^#(x, u, z, u)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) , f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) Consider the dependency graph 1: perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) -->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3 2: f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) -->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3 -->_1 f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) :2 3: f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) -->_1 f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) :3 -->_1 f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) :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). { perfectp^#(s(x)) -> c_1(f^#(x, s(0()), s(x), s(x))) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { f^#(s(x), 0(), z, u) -> c_2(f^#(x, u, minus(z, s(x)), u)) , f^#(s(x), s(y), z, u) -> c_3(f^#(x, u, z, u)) } 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(0) = {}, safe(s) = {1}, safe(minus) = {1, 2}, safe(f^#) = {2, 3}, safe(c_2) = {}, safe(c_3) = {} and precedence empty . Following symbols are considered recursive: {f^#} The recursion depth is 1. Further, following argument filtering is employed: pi(0) = [], pi(s) = [1], pi(minus) = [2], pi(f^#) = [1, 4], pi(c_2) = [1], pi(c_3) = [1] Usable defined function symbols are a subset of: {f^#} For your convenience, here are the satisfied ordering constraints: pi(f^#(s(x), 0(), z, u)) = f^#(s(; x), u;) > c_2(f^#(x, u;);) = pi(c_2(f^#(x, u, minus(z, s(x)), u))) pi(f^#(s(x), s(y), z, u)) = f^#(s(; x), u;) > c_3(f^#(x, u;);) = pi(c_3(f^#(x, u, z, u))) Hurray, we answered YES(O(1),O(n^1))