YES(?,O(n^1))
6.10/2.13	YES(?,O(n^1))
6.10/2.13	
6.10/2.13	We are left with following problem, upon which TcT provides the
6.10/2.13	certificate YES(?,O(n^1)).
6.10/2.13	
6.10/2.13	Strict Trs:
6.10/2.13	  { rec(rec(x)) -> sent(rec(x))
6.10/2.13	  , rec(sent(x)) -> sent(rec(x))
6.10/2.13	  , rec(no(x)) -> sent(rec(x))
6.10/2.13	  , rec(bot()) -> up(sent(bot()))
6.10/2.13	  , rec(up(x)) -> up(rec(x))
6.10/2.13	  , sent(up(x)) -> up(sent(x))
6.10/2.13	  , no(up(x)) -> up(no(x))
6.10/2.13	  , top(rec(up(x))) -> top(check(rec(x)))
6.10/2.13	  , top(sent(up(x))) -> top(check(rec(x)))
6.10/2.13	  , top(no(up(x))) -> top(check(rec(x)))
6.10/2.13	  , check(rec(x)) -> rec(check(x))
6.10/2.13	  , check(sent(x)) -> sent(check(x))
6.10/2.13	  , check(no(x)) -> no(x)
6.10/2.13	  , check(no(x)) -> no(check(x))
6.10/2.13	  , check(up(x)) -> up(check(x)) }
6.10/2.13	Obligation:
6.10/2.13	  innermost runtime complexity
6.10/2.13	Answer:
6.10/2.13	  YES(?,O(n^1))
6.10/2.13	
6.10/2.13	Arguments of following rules are not normal-forms:
6.10/2.13	
6.10/2.13	{ top(rec(up(x))) -> top(check(rec(x)))
6.10/2.13	, top(sent(up(x))) -> top(check(rec(x)))
6.10/2.13	, top(no(up(x))) -> top(check(rec(x))) }
6.10/2.13	
6.10/2.13	All above mentioned rules can be savely removed.
6.10/2.13	
6.10/2.13	We are left with following problem, upon which TcT provides the
6.10/2.13	certificate YES(?,O(n^1)).
6.10/2.13	
6.10/2.13	Strict Trs:
6.10/2.13	  { rec(rec(x)) -> sent(rec(x))
6.10/2.13	  , rec(sent(x)) -> sent(rec(x))
6.10/2.13	  , rec(no(x)) -> sent(rec(x))
6.10/2.13	  , rec(bot()) -> up(sent(bot()))
6.10/2.13	  , rec(up(x)) -> up(rec(x))
6.10/2.13	  , sent(up(x)) -> up(sent(x))
6.10/2.13	  , no(up(x)) -> up(no(x))
6.10/2.13	  , check(rec(x)) -> rec(check(x))
6.10/2.13	  , check(sent(x)) -> sent(check(x))
6.10/2.13	  , check(no(x)) -> no(x)
6.10/2.13	  , check(no(x)) -> no(check(x))
6.10/2.13	  , check(up(x)) -> up(check(x)) }
6.10/2.13	Obligation:
6.10/2.13	  innermost runtime complexity
6.10/2.13	Answer:
6.10/2.13	  YES(?,O(n^1))
6.10/2.13	
6.10/2.13	The problem is match-bounded by 1. The enriched problem is
6.10/2.13	compatible with the following automaton.
6.10/2.13	{ rec_0(2) -> 1
6.10/2.13	, rec_1(2) -> 3
6.10/2.13	, sent_0(2) -> 1
6.10/2.13	, sent_1(2) -> 3
6.10/2.13	, sent_1(4) -> 3
6.10/2.13	, no_0(2) -> 1
6.10/2.13	, no_1(2) -> 3
6.10/2.13	, bot_0() -> 2
6.10/2.13	, bot_1() -> 4
6.10/2.13	, up_0(2) -> 2
6.10/2.13	, up_1(3) -> 1
6.10/2.13	, up_1(3) -> 3
6.10/2.13	, check_0(2) -> 1
6.10/2.13	, check_1(2) -> 3 }
6.10/2.13	
6.10/2.13	Hurray, we answered YES(?,O(n^1))
6.10/2.14	EOF