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(g(x), y, y) -> g(f(x, x, y)) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { f^#(g(x), y, y) -> c_1(f^#(x, 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^#(g(x), y, y) -> c_1(f^#(x, x, y)) } Strict Trs: { f(g(x), y, y) -> g(f(x, x, y)) } Obligation: 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^#(g(x), y, y) -> c_1(f^#(x, x, y)) } Obligation: 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} TcT has computed the following constructor-restricted matrix interpretation. [g](x1) = [1 0] x1 + [2] [0 1] [1] [f^#](x1, x2, x3) = [1 0] x1 + [2 2] x3 + [2] [2 2] [2 2] [1] [c_1](x1) = [1 0] x1 + [1] [0 1] [1] The following symbols are considered usable {f^#} The order satisfies the following ordering constraints: [f^#(g(x), y, y)] = [1 0] x + [2 2] y + [4] [2 2] [2 2] [7] > [1 0] x + [2 2] y + [3] [2 2] [2 2] [2] = [c_1(f^#(x, x, y))] 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)). Weak DPs: { f^#(g(x), y, y) -> c_1(f^#(x, x, y)) } Obligation: 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. { f^#(g(x), y, y) -> c_1(f^#(x, x, y)) } We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Rules: Empty Obligation: runtime complexity Answer: YES(?,O(n^1)) We employ 'linear path analysis' using the following approximated dependency graph: empty Hurray, we answered YES(O(1),O(n^1))