YES(?,O(n^1))
0.00/0.83	YES(?,O(n^1))
0.00/0.83	
0.00/0.83	We are left with following problem, upon which TcT provides the
0.00/0.83	certificate YES(?,O(n^1)).
0.00/0.83	
0.00/0.83	Strict Trs:
0.00/0.83	  { a__f(X) -> f(X)
0.00/0.83	  , a__f(f(X)) -> a__c(f(g(f(X))))
0.00/0.83	  , a__c(X) -> d(X)
0.00/0.83	  , a__c(X) -> c(X)
0.00/0.83	  , a__h(X) -> a__c(d(X))
0.00/0.83	  , a__h(X) -> h(X)
0.00/0.83	  , mark(f(X)) -> a__f(mark(X))
0.00/0.83	  , mark(g(X)) -> g(X)
0.00/0.83	  , mark(d(X)) -> d(X)
0.00/0.83	  , mark(c(X)) -> a__c(X)
0.00/0.83	  , mark(h(X)) -> a__h(mark(X)) }
0.00/0.83	Obligation:
0.00/0.83	  runtime complexity
0.00/0.83	Answer:
0.00/0.83	  YES(?,O(n^1))
0.00/0.83	
0.00/0.83	The input is overlay and right-linear. Switching to innermost
0.00/0.83	rewriting.
0.00/0.83	
0.00/0.83	We are left with following problem, upon which TcT provides the
0.00/0.83	certificate YES(?,O(n^1)).
0.00/0.83	
0.00/0.83	Strict Trs:
0.00/0.83	  { a__f(X) -> f(X)
0.00/0.83	  , a__f(f(X)) -> a__c(f(g(f(X))))
0.00/0.83	  , a__c(X) -> d(X)
0.00/0.83	  , a__c(X) -> c(X)
0.00/0.83	  , a__h(X) -> a__c(d(X))
0.00/0.83	  , a__h(X) -> h(X)
0.00/0.83	  , mark(f(X)) -> a__f(mark(X))
0.00/0.83	  , mark(g(X)) -> g(X)
0.00/0.83	  , mark(d(X)) -> d(X)
0.00/0.83	  , mark(c(X)) -> a__c(X)
0.00/0.83	  , mark(h(X)) -> a__h(mark(X)) }
0.00/0.83	Obligation:
0.00/0.83	  innermost runtime complexity
0.00/0.83	Answer:
0.00/0.83	  YES(?,O(n^1))
0.00/0.83	
0.00/0.83	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.83	Order (PS,1-bounded)' as induced by the safe mapping
0.00/0.83	
0.00/0.83	 safe(a__f) = {1}, safe(f) = {1}, safe(a__c) = {1}, safe(g) = {1},
0.00/0.83	 safe(d) = {1}, safe(a__h) = {1}, safe(mark) = {}, safe(c) = {1},
0.00/0.83	 safe(h) = {1}
0.00/0.83	
0.00/0.83	and precedence
0.00/0.83	
0.00/0.83	 a__f > a__c, a__h > a__f, a__h > a__c, mark > a__f, mark > a__c,
0.00/0.83	 mark > a__h .
0.00/0.83	
0.00/0.83	Following symbols are considered recursive:
0.00/0.83	
0.00/0.83	 {mark}
0.00/0.83	
0.00/0.83	The recursion depth is 1.
0.00/0.83	
0.00/0.83	For your convenience, here are the satisfied ordering constraints:
0.00/0.83	
0.00/0.83	       a__f(; X) > f(; X)                  
0.00/0.83	                                           
0.00/0.83	  a__f(; f(; X)) > a__c(; f(; g(; f(; X))))
0.00/0.83	                                           
0.00/0.83	       a__c(; X) > d(; X)                  
0.00/0.83	                                           
0.00/0.83	       a__c(; X) > c(; X)                  
0.00/0.83	                                           
0.00/0.83	       a__h(; X) > a__c(; d(; X))          
0.00/0.83	                                           
0.00/0.83	       a__h(; X) > h(; X)                  
0.00/0.83	                                           
0.00/0.83	   mark(f(; X);) > a__f(; mark(X;))        
0.00/0.83	                                           
0.00/0.83	   mark(g(; X);) > g(; X)                  
0.00/0.83	                                           
0.00/0.83	   mark(d(; X);) > d(; X)                  
0.00/0.83	                                           
0.00/0.83	   mark(c(; X);) > a__c(; X)               
0.00/0.83	                                           
0.00/0.83	   mark(h(; X);) > a__h(; mark(X;))        
0.00/0.83	                                           
0.00/0.83	
0.00/0.83	Hurray, we answered YES(?,O(n^1))
0.00/0.84	EOF