YES(?,O(n^1))
0.00/0.42	YES(?,O(n^1))
0.00/0.42	
0.00/0.42	We are left with following problem, upon which TcT provides the
0.00/0.42	certificate YES(?,O(n^1)).
0.00/0.42	
0.00/0.42	Strict Trs:
0.00/0.42	  { f(X) -> n__f(X)
0.00/0.42	  , f(0()) -> cons(0(), n__f(s(0())))
0.00/0.42	  , f(s(0())) -> f(p(s(0())))
0.00/0.42	  , p(s(0())) -> 0()
0.00/0.42	  , activate(X) -> X
0.00/0.42	  , activate(n__f(X)) -> f(X) }
0.00/0.42	Obligation:
0.00/0.42	  runtime complexity
0.00/0.42	Answer:
0.00/0.42	  YES(?,O(n^1))
0.00/0.42	
0.00/0.42	The input is overlay and right-linear. Switching to innermost
0.00/0.42	rewriting.
0.00/0.42	
0.00/0.42	We are left with following problem, upon which TcT provides the
0.00/0.42	certificate YES(?,O(n^1)).
0.00/0.42	
0.00/0.42	Strict Trs:
0.00/0.42	  { f(X) -> n__f(X)
0.00/0.42	  , f(0()) -> cons(0(), n__f(s(0())))
0.00/0.42	  , f(s(0())) -> f(p(s(0())))
0.00/0.42	  , p(s(0())) -> 0()
0.00/0.42	  , activate(X) -> X
0.00/0.42	  , activate(n__f(X)) -> f(X) }
0.00/0.42	Obligation:
0.00/0.42	  innermost runtime complexity
0.00/0.42	Answer:
0.00/0.42	  YES(?,O(n^1))
0.00/0.42	
0.00/0.42	The problem is match-bounded by 2. The enriched problem is
0.00/0.42	compatible with the following automaton.
0.00/0.42	{ f_0(2) -> 1
0.00/0.42	, f_1(2) -> 1
0.00/0.42	, f_1(7) -> 1
0.00/0.42	, 0_0() -> 1
0.00/0.42	, 0_0() -> 2
0.00/0.42	, 0_1() -> 1
0.00/0.42	, 0_1() -> 3
0.00/0.42	, 0_1() -> 6
0.00/0.42	, 0_2() -> 7
0.00/0.42	, cons_0(2, 2) -> 1
0.00/0.42	, cons_0(2, 2) -> 2
0.00/0.42	, cons_1(3, 4) -> 1
0.00/0.42	, cons_2(7, 8) -> 1
0.00/0.42	, n__f_0(2) -> 1
0.00/0.42	, n__f_0(2) -> 2
0.00/0.42	, n__f_1(2) -> 1
0.00/0.42	, n__f_1(5) -> 4
0.00/0.42	, n__f_2(2) -> 1
0.00/0.42	, n__f_2(7) -> 1
0.00/0.42	, n__f_2(9) -> 8
0.00/0.42	, s_0(2) -> 1
0.00/0.42	, s_0(2) -> 2
0.00/0.42	, s_1(6) -> 5
0.00/0.42	, s_2(7) -> 9
0.00/0.42	, p_0(2) -> 1
0.00/0.42	, p_1(5) -> 7
0.00/0.42	, activate_0(2) -> 1 }
0.00/0.42	
0.00/0.42	Hurray, we answered YES(?,O(n^1))
0.00/0.42	EOF