YES(?,O(n^1)) 4.87/1.30 YES(?,O(n^1)) 4.87/1.30 4.87/1.30 We are left with following problem, upon which TcT provides the 4.87/1.30 certificate YES(?,O(n^1)). 4.87/1.30 4.87/1.30 Strict Trs: 4.87/1.30 { a__zeros() -> cons(0(), zeros()) 4.87/1.30 , a__zeros() -> zeros() 4.87/1.30 , a__tail(X) -> tail(X) 4.87/1.30 , a__tail(cons(X, XS)) -> mark(XS) 4.87/1.30 , mark(cons(X1, X2)) -> cons(mark(X1), X2) 4.87/1.30 , mark(0()) -> 0() 4.87/1.30 , mark(zeros()) -> a__zeros() 4.87/1.30 , mark(tail(X)) -> a__tail(mark(X)) } 4.87/1.30 Obligation: 4.87/1.30 runtime complexity 4.87/1.30 Answer: 4.87/1.30 YES(?,O(n^1)) 4.87/1.30 4.87/1.30 The input is overlay and right-linear. Switching to innermost 4.87/1.30 rewriting. 4.87/1.30 4.87/1.30 We are left with following problem, upon which TcT provides the 4.87/1.30 certificate YES(?,O(n^1)). 4.87/1.30 4.87/1.30 Strict Trs: 4.87/1.30 { a__zeros() -> cons(0(), zeros()) 4.87/1.30 , a__zeros() -> zeros() 4.87/1.30 , a__tail(X) -> tail(X) 4.87/1.30 , a__tail(cons(X, XS)) -> mark(XS) 4.87/1.30 , mark(cons(X1, X2)) -> cons(mark(X1), X2) 4.87/1.30 , mark(0()) -> 0() 4.87/1.30 , mark(zeros()) -> a__zeros() 4.87/1.30 , mark(tail(X)) -> a__tail(mark(X)) } 4.87/1.30 Obligation: 4.87/1.30 innermost runtime complexity 4.87/1.30 Answer: 4.87/1.30 YES(?,O(n^1)) 4.87/1.30 4.87/1.30 The problem is match-bounded by 4. The enriched problem is 4.87/1.30 compatible with the following automaton. 4.87/1.30 { a__zeros_0() -> 1 4.87/1.30 , a__zeros_1() -> 1 4.87/1.30 , a__zeros_2() -> 1 4.87/1.30 , a__zeros_3() -> 1 4.87/1.30 , cons_0(2, 2) -> 2 4.87/1.30 , cons_1(1, 2) -> 1 4.87/1.30 , cons_1(3, 4) -> 1 4.87/1.30 , cons_2(5, 6) -> 1 4.87/1.30 , cons_3(7, 8) -> 1 4.87/1.30 , cons_4(9, 10) -> 1 4.87/1.30 , 0_0() -> 2 4.87/1.30 , 0_1() -> 1 4.87/1.30 , 0_1() -> 3 4.87/1.30 , 0_2() -> 5 4.87/1.30 , 0_3() -> 7 4.87/1.30 , 0_4() -> 9 4.87/1.30 , zeros_0() -> 2 4.87/1.30 , zeros_1() -> 1 4.87/1.30 , zeros_1() -> 4 4.87/1.30 , zeros_2() -> 1 4.87/1.30 , zeros_2() -> 6 4.87/1.30 , zeros_3() -> 1 4.87/1.30 , zeros_3() -> 8 4.87/1.30 , zeros_4() -> 1 4.87/1.30 , zeros_4() -> 10 4.87/1.30 , a__tail_0(2) -> 1 4.87/1.30 , a__tail_1(1) -> 1 4.87/1.30 , mark_0(2) -> 1 4.87/1.30 , mark_1(2) -> 1 4.87/1.30 , mark_2(2) -> 1 4.87/1.30 , mark_2(4) -> 1 4.87/1.30 , mark_2(6) -> 1 4.87/1.30 , mark_2(8) -> 1 4.87/1.30 , mark_2(10) -> 1 4.87/1.30 , tail_0(2) -> 2 4.87/1.30 , tail_1(2) -> 1 4.87/1.30 , tail_2(1) -> 1 } 4.87/1.30 4.87/1.30 Hurray, we answered YES(?,O(n^1)) 4.87/1.30 EOF