YES(?,O(n^1))
0.00/0.59	YES(?,O(n^1))
0.00/0.59	
0.00/0.59	We are left with following problem, upon which TcT provides the
0.00/0.59	certificate YES(?,O(n^1)).
0.00/0.59	
0.00/0.59	Strict Trs:
0.00/0.59	  { f(a, empty()) -> g(a, empty())
0.00/0.59	  , f(a, cons(x, k)) -> f(cons(x, a), k)
0.00/0.59	  , g(empty(), d) -> d
0.00/0.59	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
0.00/0.59	Obligation:
0.00/0.59	  runtime complexity
0.00/0.59	Answer:
0.00/0.59	  YES(?,O(n^1))
0.00/0.59	
0.00/0.59	The input is overlay and right-linear. Switching to innermost
0.00/0.59	rewriting.
0.00/0.59	
0.00/0.59	We are left with following problem, upon which TcT provides the
0.00/0.59	certificate YES(?,O(n^1)).
0.00/0.59	
0.00/0.59	Strict Trs:
0.00/0.59	  { f(a, empty()) -> g(a, empty())
0.00/0.59	  , f(a, cons(x, k)) -> f(cons(x, a), k)
0.00/0.59	  , g(empty(), d) -> d
0.00/0.59	  , g(cons(x, k), d) -> g(k, cons(x, d)) }
0.00/0.59	Obligation:
0.00/0.59	  innermost runtime complexity
0.00/0.59	Answer:
0.00/0.59	  YES(?,O(n^1))
0.00/0.59	
0.00/0.59	The problem is match-bounded by 2. The enriched problem is
0.00/0.59	compatible with the following automaton.
0.00/0.59	{ f_0(2, 2) -> 1
0.00/0.59	, f_1(4, 2) -> 1
0.00/0.59	, empty_0() -> 1
0.00/0.59	, empty_0() -> 2
0.00/0.59	, empty_1() -> 1
0.00/0.59	, empty_1() -> 3
0.00/0.59	, g_0(2, 2) -> 1
0.00/0.59	, g_1(2, 3) -> 1
0.00/0.59	, g_1(2, 4) -> 1
0.00/0.59	, g_1(4, 3) -> 1
0.00/0.59	, g_2(2, 5) -> 1
0.00/0.59	, g_2(4, 5) -> 1
0.00/0.59	, cons_0(2, 2) -> 1
0.00/0.59	, cons_0(2, 2) -> 2
0.00/0.59	, cons_1(2, 2) -> 1
0.00/0.59	, cons_1(2, 2) -> 4
0.00/0.59	, cons_1(2, 3) -> 1
0.00/0.59	, cons_1(2, 3) -> 3
0.00/0.59	, cons_1(2, 4) -> 1
0.00/0.59	, cons_1(2, 4) -> 4
0.00/0.59	, cons_1(2, 5) -> 1
0.00/0.59	, cons_1(2, 5) -> 3
0.00/0.59	, cons_2(2, 3) -> 1
0.00/0.59	, cons_2(2, 3) -> 5
0.00/0.59	, cons_2(2, 5) -> 1
0.00/0.59	, cons_2(2, 5) -> 5 }
0.00/0.59	
0.00/0.59	Hurray, we answered YES(?,O(n^1))
0.00/0.59	EOF