We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { a(b(x)) -> a(c(b(x))) } Obligation: runtime complexity Answer: YES(O(1),O(1)) The input is overlay and right-linear. Switching to innermost rewriting. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict Trs: { a(b(x)) -> a(c(b(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) We add the following weak dependency pairs: Strict DPs: { a^#(b(x)) -> c_1(a^#(c(b(x)))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Strict DPs: { a^#(b(x)) -> c_1(a^#(c(b(x)))) } Strict Trs: { a(b(x)) -> a(c(b(x))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(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(1)). Strict DPs: { a^#(b(x)) -> c_1(a^#(c(b(x)))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: none TcT has computed the following constructor-restricted matrix interpretation. [b](x1) = [0] [2] [c](x1) = [0] [0] [a^#](x1) = [0 2] x1 + [0] [0 0] [0] [c_1](x1) = [0] [0] The order satisfies the following ordering constraints: [a^#(b(x))] = [4] [0] > [0] [0] = [c_1(a^#(c(b(x))))] 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(1)). Weak DPs: { a^#(b(x)) -> c_1(a^#(c(b(x)))) } 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. { a^#(b(x)) -> c_1(a^#(c(b(x)))) } 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(1))