YES(?,O(1)) We are left with following problem, upon which TcT provides the certificate YES(?,O(1)). Strict Trs: { f(X) -> n__f(X) , f(f(X)) -> c(n__f(g(n__f(X)))) , c(X) -> d(activate(X)) , d(X) -> n__d(X) , activate(X) -> X , activate(n__f(X)) -> f(X) , activate(n__d(X)) -> d(X) , h(X) -> c(n__d(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(1)) Arguments of following rules are not normal-forms: { f(f(X)) -> c(n__f(g(n__f(X)))) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(?,O(1)). Strict Trs: { f(X) -> n__f(X) , c(X) -> d(activate(X)) , d(X) -> n__d(X) , activate(X) -> X , activate(n__f(X)) -> f(X) , activate(n__d(X)) -> d(X) , h(X) -> c(n__d(X)) } Obligation: innermost runtime complexity Answer: YES(?,O(1)) The input was oriented with the instance of 'Small Polynomial Path Order (PS,0-bounded)' as induced by the safe mapping safe(f) = {1}, safe(c) = {1}, safe(n__f) = {1}, safe(g) = {1}, safe(d) = {1}, safe(activate) = {1}, safe(h) = {1}, safe(n__d) = {1} and precedence c > f, c > d, c > activate, activate > f, activate > d, h > f, h > c, h > d, h > activate, f ~ d . Following symbols are considered recursive: {} The recursion depth is 0. For your convenience, here are the satisfied ordering constraints: f(; X) > n__f(; X) c(; X) > d(; activate(; X)) d(; X) > n__d(; X) activate(; X) > X activate(; n__f(; X)) > f(; X) activate(; n__d(; X)) > d(; X) h(; X) > c(; n__d(; X)) Hurray, we answered YES(?,O(1))