We are left with following problem, upon which TcT provides the certificate YES(?,O(n^1)). Strict Trs: { f(X) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X , if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(X) } Obligation: runtime complexity Answer: YES(?,O(n^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(n^1)). Strict Trs: { f(X) -> if(X, c(), n__f(true())) , f(X) -> n__f(X) , if(true(), X, Y) -> X , if(false(), X, Y) -> activate(Y) , activate(X) -> X , activate(n__f(X)) -> f(X) } Obligation: innermost runtime complexity Answer: YES(?,O(n^1)) The problem is match-bounded by 3. The enriched problem is compatible with the following automaton. { f_0(3) -> 1 , f_0(4) -> 1 , f_0(5) -> 1 , f_0(6) -> 1 , f_1(3) -> 2 , f_1(3) -> 7 , f_1(4) -> 2 , f_1(4) -> 7 , f_1(5) -> 2 , f_1(5) -> 7 , f_1(6) -> 2 , f_1(6) -> 7 , f_2(10) -> 1 , f_2(13) -> 2 , f_2(13) -> 7 , if_0(3, 3, 3) -> 2 , if_0(3, 3, 4) -> 2 , if_0(3, 3, 5) -> 2 , if_0(3, 3, 6) -> 2 , if_0(3, 4, 3) -> 2 , if_0(3, 4, 4) -> 2 , if_0(3, 4, 5) -> 2 , if_0(3, 4, 6) -> 2 , if_0(3, 5, 3) -> 2 , if_0(3, 5, 4) -> 2 , if_0(3, 5, 5) -> 2 , if_0(3, 5, 6) -> 2 , if_0(3, 6, 3) -> 2 , if_0(3, 6, 4) -> 2 , if_0(3, 6, 5) -> 2 , if_0(3, 6, 6) -> 2 , if_0(4, 3, 3) -> 2 , if_0(4, 3, 4) -> 2 , if_0(4, 3, 5) -> 2 , if_0(4, 3, 6) -> 2 , if_0(4, 4, 3) -> 2 , if_0(4, 4, 4) -> 2 , if_0(4, 4, 5) -> 2 , if_0(4, 4, 6) -> 2 , if_0(4, 5, 3) -> 2 , if_0(4, 5, 4) -> 2 , if_0(4, 5, 5) -> 2 , if_0(4, 5, 6) -> 2 , if_0(4, 6, 3) -> 2 , if_0(4, 6, 4) -> 2 , if_0(4, 6, 5) -> 2 , if_0(4, 6, 6) -> 2 , if_0(5, 3, 3) -> 2 , if_0(5, 3, 4) -> 2 , if_0(5, 3, 5) -> 2 , if_0(5, 3, 6) -> 2 , if_0(5, 4, 3) -> 2 , if_0(5, 4, 4) -> 2 , if_0(5, 4, 5) -> 2 , if_0(5, 4, 6) -> 2 , if_0(5, 5, 3) -> 2 , if_0(5, 5, 4) -> 2 , if_0(5, 5, 5) -> 2 , if_0(5, 5, 6) -> 2 , if_0(5, 6, 3) -> 2 , if_0(5, 6, 4) -> 2 , if_0(5, 6, 5) -> 2 , if_0(5, 6, 6) -> 2 , if_0(6, 3, 3) -> 2 , if_0(6, 3, 4) -> 2 , if_0(6, 3, 5) -> 2 , if_0(6, 3, 6) -> 2 , if_0(6, 4, 3) -> 2 , if_0(6, 4, 4) -> 2 , if_0(6, 4, 5) -> 2 , if_0(6, 4, 6) -> 2 , if_0(6, 5, 3) -> 2 , if_0(6, 5, 4) -> 2 , if_0(6, 5, 5) -> 2 , if_0(6, 5, 6) -> 2 , if_0(6, 6, 3) -> 2 , if_0(6, 6, 4) -> 2 , if_0(6, 6, 5) -> 2 , if_0(6, 6, 6) -> 2 , if_1(3, 8, 9) -> 1 , if_1(4, 8, 9) -> 1 , if_1(5, 8, 9) -> 1 , if_1(6, 8, 9) -> 1 , if_2(3, 11, 12) -> 2 , if_2(3, 11, 12) -> 7 , if_2(4, 11, 12) -> 2 , if_2(4, 11, 12) -> 7 , if_2(5, 11, 12) -> 2 , if_2(5, 11, 12) -> 7 , if_2(6, 11, 12) -> 2 , if_2(6, 11, 12) -> 7 , if_3(10, 14, 15) -> 1 , if_3(13, 14, 15) -> 2 , if_3(13, 14, 15) -> 7 , c_0() -> 2 , c_0() -> 3 , c_0() -> 7 , c_1() -> 1 , c_1() -> 8 , c_2() -> 2 , c_2() -> 7 , c_2() -> 11 , c_3() -> 1 , c_3() -> 2 , c_3() -> 7 , c_3() -> 14 , n__f_0(3) -> 2 , n__f_0(3) -> 4 , n__f_0(3) -> 7 , n__f_0(4) -> 2 , n__f_0(4) -> 4 , n__f_0(4) -> 7 , n__f_0(5) -> 2 , n__f_0(5) -> 4 , n__f_0(5) -> 7 , n__f_0(6) -> 2 , n__f_0(6) -> 4 , n__f_0(6) -> 7 , n__f_1(3) -> 1 , n__f_1(4) -> 1 , n__f_1(5) -> 1 , n__f_1(6) -> 1 , n__f_1(10) -> 1 , n__f_1(10) -> 9 , n__f_2(3) -> 2 , n__f_2(3) -> 7 , n__f_2(4) -> 2 , n__f_2(4) -> 7 , n__f_2(5) -> 2 , n__f_2(5) -> 7 , n__f_2(6) -> 2 , n__f_2(6) -> 7 , n__f_2(13) -> 2 , n__f_2(13) -> 7 , n__f_2(13) -> 12 , n__f_3(10) -> 1 , n__f_3(13) -> 2 , n__f_3(13) -> 7 , n__f_3(16) -> 15 , true_0() -> 2 , true_0() -> 5 , true_0() -> 7 , true_1() -> 10 , true_2() -> 13 , true_3() -> 16 , false_0() -> 2 , false_0() -> 6 , false_0() -> 7 , activate_0(3) -> 7 , activate_0(4) -> 7 , activate_0(5) -> 7 , activate_0(6) -> 7 , activate_1(3) -> 2 , activate_1(4) -> 2 , activate_1(5) -> 2 , activate_1(6) -> 2 , activate_1(9) -> 1 , activate_1(12) -> 2 , activate_1(12) -> 7 } Hurray, we answered YES(?,O(n^1))