YES(?,O(1))
0.00/0.22	YES(?,O(1))
0.00/0.22	
0.00/0.22	We are left with following problem, upon which TcT provides the
0.00/0.22	certificate YES(?,O(1)).
0.00/0.22	
0.00/0.22	Strict Trs:
0.00/0.22	  { app(nil(), YS) -> YS
0.00/0.22	  , app(cons(X), YS) -> cons(X)
0.00/0.22	  , from(X) -> cons(X)
0.00/0.22	  , zWadr(XS, nil()) -> nil()
0.00/0.22	  , zWadr(nil(), YS) -> nil()
0.00/0.22	  , zWadr(cons(X), cons(Y)) -> cons(app(Y, cons(X)))
0.00/0.22	  , prefix(L) -> cons(nil()) }
0.00/0.22	Obligation:
0.00/0.22	  runtime complexity
0.00/0.22	Answer:
0.00/0.22	  YES(?,O(1))
0.00/0.22	
0.00/0.22	The input is overlay and right-linear. Switching to innermost
0.00/0.22	rewriting.
0.00/0.22	
0.00/0.22	We are left with following problem, upon which TcT provides the
0.00/0.22	certificate YES(?,O(1)).
0.00/0.22	
0.00/0.22	Strict Trs:
0.00/0.22	  { app(nil(), YS) -> YS
0.00/0.22	  , app(cons(X), YS) -> cons(X)
0.00/0.22	  , from(X) -> cons(X)
0.00/0.22	  , zWadr(XS, nil()) -> nil()
0.00/0.22	  , zWadr(nil(), YS) -> nil()
0.00/0.22	  , zWadr(cons(X), cons(Y)) -> cons(app(Y, cons(X)))
0.00/0.22	  , prefix(L) -> cons(nil()) }
0.00/0.22	Obligation:
0.00/0.22	  innermost runtime complexity
0.00/0.22	Answer:
0.00/0.22	  YES(?,O(1))
0.00/0.22	
0.00/0.22	The input was oriented with the instance of 'Small Polynomial Path
0.00/0.22	Order (PS,0-bounded)' as induced by the safe mapping
0.00/0.22	
0.00/0.22	 safe(app) = {1, 2}, safe(nil) = {}, safe(cons) = {1},
0.00/0.22	 safe(from) = {}, safe(zWadr) = {}, safe(prefix) = {1}
0.00/0.22	
0.00/0.22	and precedence
0.00/0.22	
0.00/0.22	 app > from, zWadr > app, zWadr > from, zWadr > prefix,
0.00/0.22	 prefix > from, app ~ prefix .
0.00/0.22	
0.00/0.22	Following symbols are considered recursive:
0.00/0.22	
0.00/0.22	 {}
0.00/0.22	
0.00/0.22	The recursion depth is 0.
0.00/0.22	
0.00/0.22	For your convenience, here are the satisfied ordering constraints:
0.00/0.22	
0.00/0.22	              app(; nil(),  YS) > YS                          
0.00/0.22	                                                              
0.00/0.22	          app(; cons(; X),  YS) > cons(; X)                   
0.00/0.22	                                                              
0.00/0.22	                       from(X;) > cons(; X)                   
0.00/0.22	                                                              
0.00/0.22	             zWadr(XS,  nil();) > nil()                       
0.00/0.22	                                                              
0.00/0.22	             zWadr(nil(),  YS;) > nil()                       
0.00/0.22	                                                              
0.00/0.22	  zWadr(cons(; X),  cons(; Y);) > cons(; app(; Y,  cons(; X)))
0.00/0.22	                                                              
0.00/0.22	                    prefix(; L) > cons(; nil())               
0.00/0.22	                                                              
0.00/0.22	
0.00/0.22	Hurray, we answered YES(?,O(1))
0.00/0.22	EOF