YES(?,O(n^1))
0.00/0.99	YES(?,O(n^1))
0.00/0.99	
0.00/0.99	We are left with following problem, upon which TcT provides the
0.00/0.99	certificate YES(?,O(n^1)).
0.00/0.99	
0.00/0.99	Strict Trs:
0.00/0.99	  { concat(leaf(), Y) -> Y
0.00/0.99	  , concat(cons(U, V), Y) -> cons(U, concat(V, Y))
0.00/0.99	  , lessleaves(X, leaf()) -> false()
0.00/0.99	  , lessleaves(leaf(), cons(W, Z)) -> true()
0.00/0.99	  , lessleaves(cons(U, V), cons(W, Z)) ->
0.00/0.99	    lessleaves(concat(U, V), concat(W, Z)) }
0.00/0.99	Obligation:
0.00/0.99	  innermost runtime complexity
0.00/0.99	Answer:
0.00/0.99	  YES(?,O(n^1))
0.00/0.99	
0.00/0.99	The problem is match-bounded by 2. The enriched problem is
0.00/0.99	compatible with the following automaton.
0.00/0.99	{ concat_0(2, 2) -> 1
0.00/0.99	, concat_1(2, 2) -> 3
0.00/0.99	, concat_1(2, 2) -> 4
0.00/0.99	, concat_1(2, 2) -> 5
0.00/0.99	, concat_1(2, 3) -> 3
0.00/0.99	, concat_1(2, 3) -> 4
0.00/0.99	, concat_1(2, 3) -> 5
0.00/0.99	, concat_2(2, 3) -> 4
0.00/0.99	, concat_2(2, 3) -> 5
0.00/0.99	, leaf_0() -> 1
0.00/0.99	, leaf_0() -> 2
0.00/0.99	, leaf_0() -> 3
0.00/0.99	, leaf_0() -> 4
0.00/0.99	, leaf_0() -> 5
0.00/0.99	, cons_0(2, 2) -> 1
0.00/0.99	, cons_0(2, 2) -> 2
0.00/0.99	, cons_0(2, 2) -> 3
0.00/0.99	, cons_0(2, 2) -> 4
0.00/0.99	, cons_0(2, 2) -> 5
0.00/0.99	, cons_1(2, 3) -> 1
0.00/0.99	, cons_1(2, 3) -> 3
0.00/0.99	, cons_1(2, 3) -> 4
0.00/0.99	, cons_1(2, 3) -> 5
0.00/0.99	, lessleaves_0(2, 2) -> 1
0.00/0.99	, lessleaves_1(3, 3) -> 1
0.00/0.99	, lessleaves_2(4, 5) -> 1
0.00/0.99	, false_0() -> 1
0.00/0.99	, false_0() -> 2
0.00/0.99	, false_0() -> 3
0.00/0.99	, false_0() -> 4
0.00/0.99	, false_0() -> 5
0.00/0.99	, false_1() -> 1
0.00/0.99	, true_0() -> 1
0.00/0.99	, true_0() -> 2
0.00/0.99	, true_0() -> 3
0.00/0.99	, true_0() -> 4
0.00/0.99	, true_0() -> 5
0.00/0.99	, true_1() -> 1 }
0.00/0.99	
0.00/0.99	Hurray, we answered YES(?,O(n^1))
0.00/1.00	EOF