YES(?,O(n^1))
2.97/1.01	YES(?,O(n^1))
2.97/1.01	
2.97/1.01	We are left with following problem, upon which TcT provides the
2.97/1.01	certificate YES(?,O(n^1)).
2.97/1.01	
2.97/1.01	Strict Trs:
2.97/1.01	  { f(X) -> n__f(X)
2.97/1.01	  , f(f(X)) -> c(n__f(g(n__f(X))))
2.97/1.01	  , c(X) -> d(activate(X))
2.97/1.01	  , d(X) -> n__d(X)
2.97/1.01	  , activate(X) -> X
2.97/1.01	  , activate(n__f(X)) -> f(X)
2.97/1.01	  , activate(n__d(X)) -> d(X)
2.97/1.01	  , h(X) -> c(n__d(X)) }
2.97/1.01	Obligation:
2.97/1.01	  runtime complexity
2.97/1.01	Answer:
2.97/1.01	  YES(?,O(n^1))
2.97/1.01	
2.97/1.01	The problem is match-bounded by 4. The enriched problem is
2.97/1.01	compatible with the following automaton.
2.97/1.01	{ f_0(2) -> 1
2.97/1.01	, f_0(2) -> 4
2.97/1.01	, f_1(2) -> 1
2.97/1.01	, f_1(2) -> 3
2.97/1.01	, f_1(2) -> 4
2.97/1.01	, f_2(2) -> 4
2.97/1.01	, f_3(2) -> 4
2.97/1.01	, c_0(2) -> 1
2.97/1.01	, c_0(2) -> 4
2.97/1.01	, c_1(1) -> 1
2.97/1.01	, c_1(1) -> 4
2.97/1.01	, n__f_0(2) -> 1
2.97/1.01	, n__f_0(2) -> 2
2.97/1.01	, n__f_0(2) -> 3
2.97/1.01	, n__f_0(2) -> 4
2.97/1.01	, n__f_1(2) -> 1
2.97/1.01	, n__f_1(2) -> 4
2.97/1.01	, n__f_2(2) -> 1
2.97/1.01	, n__f_2(2) -> 3
2.97/1.01	, n__f_2(2) -> 4
2.97/1.01	, n__f_3(2) -> 4
2.97/1.01	, n__f_4(2) -> 4
2.97/1.01	, g_0(2) -> 1
2.97/1.01	, g_0(2) -> 2
2.97/1.01	, g_0(2) -> 3
2.97/1.01	, g_0(2) -> 4
2.97/1.01	, d_0(2) -> 1
2.97/1.01	, d_0(2) -> 4
2.97/1.01	, d_1(2) -> 1
2.97/1.01	, d_1(2) -> 3
2.97/1.01	, d_1(2) -> 4
2.97/1.01	, d_1(3) -> 1
2.97/1.01	, d_1(3) -> 4
2.97/1.01	, d_2(2) -> 4
2.97/1.01	, d_2(4) -> 1
2.97/1.01	, d_2(4) -> 4
2.97/1.01	, d_3(2) -> 4
2.97/1.01	, d_3(3) -> 4
2.97/1.01	, d_3(4) -> 4
2.97/1.01	, activate_0(2) -> 1
2.97/1.01	, activate_0(2) -> 4
2.97/1.01	, activate_1(2) -> 3
2.97/1.01	, activate_2(1) -> 4
2.97/1.01	, h_0(2) -> 1
2.97/1.01	, h_0(2) -> 4
2.97/1.01	, n__d_0(2) -> 1
2.97/1.01	, n__d_0(2) -> 2
2.97/1.01	, n__d_0(2) -> 3
2.97/1.01	, n__d_0(2) -> 4
2.97/1.01	, n__d_1(2) -> 1
2.97/1.01	, n__d_1(2) -> 4
2.97/1.01	, n__d_2(2) -> 1
2.97/1.01	, n__d_2(2) -> 3
2.97/1.01	, n__d_2(2) -> 4
2.97/1.01	, n__d_2(3) -> 1
2.97/1.01	, n__d_2(3) -> 4
2.97/1.01	, n__d_3(2) -> 4
2.97/1.01	, n__d_3(4) -> 1
2.97/1.01	, n__d_3(4) -> 4
2.97/1.01	, n__d_4(2) -> 4
2.97/1.01	, n__d_4(3) -> 4
2.97/1.01	, n__d_4(4) -> 4 }
2.97/1.01	
2.97/1.01	Hurray, we answered YES(?,O(n^1))
2.97/1.01	EOF