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: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) , mark(f(X)) -> a__f(X) , mark(a()) -> a() , mark(g(X)) -> g(mark(X)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add following weak dependency pairs: Strict DPs: { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> c_4() , mark^#(g(X)) -> c_5(mark^#(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: { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> c_4() , mark^#(g(X)) -> c_5(mark^#(X)) } Strict Trs: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) , mark(f(X)) -> a__f(X) , mark(a()) -> a() , mark(g(X)) -> g(mark(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: { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> c_4() , mark^#(g(X)) -> c_5(mark^#(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_2) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1} TcT has computed following constructor-restricted matrix interpretation. [f](x1) = [1] x1 + [0] [a] = [0] [g](x1) = [0] [a__f^#](x1) = [1] [c_1] = [0] [c_2](x1) = [1] x1 + [0] [mark^#](x1) = [0] [c_3](x1) = [1] x1 + [0] [c_4] = [1] [c_5](x1) = [1] x1 + [2] 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: { a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> c_4() , mark^#(g(X)) -> c_5(mark^#(X)) } Weak DPs: { a__f^#(X) -> c_1() } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1,3} by applications of Pre({1,3}) = {2,4}. Here rules are labeled as follows: DPs: { 1: a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , 2: mark^#(f(X)) -> c_3(a__f^#(X)) , 3: mark^#(a()) -> c_4() , 4: mark^#(g(X)) -> c_5(mark^#(X)) , 5: a__f^#(X) -> c_1() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(g(X)) -> c_5(mark^#(X)) } Weak DPs: { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(a()) -> c_4() } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) We estimate the number of application of {1} by applications of Pre({1}) = {2}. Here rules are labeled as follows: DPs: { 1: mark^#(f(X)) -> c_3(a__f^#(X)) , 2: mark^#(g(X)) -> c_5(mark^#(X)) , 3: a__f^#(X) -> c_1() , 4: a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , 5: mark^#(a()) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_5(mark^#(X)) } Weak DPs: { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> 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. { a__f^#(X) -> c_1() , a__f^#(f(a())) -> c_2(a__f^#(g(f(a())))) , mark^#(f(X)) -> c_3(a__f^#(X)) , mark^#(a()) -> c_4() } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict DPs: { mark^#(g(X)) -> c_5(mark^#(X)) } 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(g) = {1}, safe(mark^#) = {}, safe(c_5) = {} and precedence empty . Following symbols are considered recursive: {mark^#} The recursion depth is 1. Further, following argument filtering is employed: pi(g) = [1], pi(mark^#) = [1], pi(c_5) = [1] Usable defined function symbols are a subset of: {mark^#} For your convenience, here are the satisfied ordering constraints: pi(mark^#(g(X))) = mark^#(g(; X);) > c_5(mark^#(X;);) = pi(c_5(mark^#(X))) Hurray, we answered YES(O(1),O(n^1))