0.00/0.03	YES
0.00/0.03	
0.00/0.03	Problem 1: 
0.00/0.03	
0.00/0.03	(VAR N X XS Y YS)
0.00/0.03	(STRATEGY CONTEXTSENSITIVE
0.00/0.03	(from 1)
0.00/0.03	(minus 1 2)
0.00/0.03	(quot 1 2)
0.00/0.03	(sel 1 2)
0.00/0.03	(zWquot 1 2)
0.00/0.03	(0)
0.00/0.03	(cons 1)
0.00/0.03	(nil)
0.00/0.03	(s 1)
0.00/0.03	)
0.00/0.03	(RULES
0.00/0.03	from(X) -> cons(X,from(s(X)))
0.00/0.03	minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	minus(X,0) -> 0
0.00/0.03	quot(0,s(Y)) -> 0
0.00/0.03	quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	sel(0,cons(X,XS)) -> X
0.00/0.03	sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	zWquot(nil,XS) -> nil
0.00/0.03	zWquot(XS,nil) -> nil
0.00/0.03	)
0.00/0.03	
0.00/0.03	Problem 1: 
0.00/0.03	
0.00/0.03	Dependency Pairs Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	 QUOT(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
0.00/0.03	 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
0.00/0.03	 SEL(s(N),cons(X,XS)) -> XS
0.00/0.03	 ZWQUOT(cons(X,XS),cons(Y,YS)) -> QUOT(X,Y)
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding Rules:
0.00/0.03	 from(s(X)) -> FROM(s(X))
0.00/0.03	 zWquot(XS,YS) -> ZWQUOT(XS,YS)
0.00/0.03	 zWquot(XS,x5) -> x5
0.00/0.03	 zWquot(x5,YS) -> x5
0.00/0.03	
0.00/0.03	Problem 1: 
0.00/0.03	
0.00/0.03	SCC Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	 QUOT(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
0.00/0.03	 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
0.00/0.03	 SEL(s(N),cons(X,XS)) -> XS
0.00/0.03	 ZWQUOT(cons(X,XS),cons(Y,YS)) -> QUOT(X,Y)
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 from(s(X)) -> FROM(s(X))
0.00/0.03	 zWquot(XS,YS) -> ZWQUOT(XS,YS)
0.00/0.03	 zWquot(XS,x5) -> x5
0.00/0.03	 zWquot(x5,YS) -> x5
0.00/0.03	->Strongly Connected Components:
0.00/0.03	->->Cycle:
0.00/0.03	->->-> Pairs:
0.00/0.03	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	->->-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	->->-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	->->Cycle:
0.00/0.03	->->-> Pairs:
0.00/0.03	 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
0.00/0.03	->->-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	->->-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	->->Cycle:
0.00/0.03	->->-> Pairs:
0.00/0.03	 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
0.00/0.03	->->-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	->->-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	
0.00/0.03	
0.00/0.03	The problem is decomposed in 3 subproblems.
0.00/0.03	
0.00/0.03	Problem 1.1: 
0.00/0.03	
0.00/0.03	SubNColl Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	->Projection:
0.00/0.03	 pi(MINUS) = 1
0.00/0.03	
0.00/0.03	Problem 1.1: 
0.00/0.03	
0.00/0.03	Basic Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 Empty
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	-> Result:
0.00/0.03	 Set P is empty
0.00/0.03	
0.00/0.03	The problem is finite.
0.00/0.03	
0.00/0.03	Problem 1.2: 
0.00/0.03	
0.00/0.03	Reduction Pairs Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 QUOT(s(X),s(Y)) -> QUOT(minus(X,Y),s(Y))
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	-> Usable rules:
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	->Interpretation type:
0.00/0.03	 Linear
0.00/0.03	->Coefficients:
0.00/0.03	 Natural Numbers
0.00/0.03	->Dimension:
0.00/0.03	 1
0.00/0.03	->Bound:
0.00/0.03	 2
0.00/0.03	->Interpretation:
0.00/0.03	 
0.00/0.03	[minus](X1,X2) = 1
0.00/0.03	[0] = 0
0.00/0.03	[s](X) = 2
0.00/0.03	[QUOT](X1,X2) = 2.X1
0.00/0.03	
0.00/0.03	Problem 1.2: 
0.00/0.03	
0.00/0.03	Basic Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 Empty
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	-> Result:
0.00/0.03	 Set P is empty
0.00/0.03	
0.00/0.03	The problem is finite.
0.00/0.03	
0.00/0.03	Problem 1.3: 
0.00/0.03	
0.00/0.03	SubNColl Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 SEL(s(N),cons(X,XS)) -> SEL(N,XS)
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	->Projection:
0.00/0.03	 pi(SEL) = 1
0.00/0.03	
0.00/0.03	Problem 1.3: 
0.00/0.03	
0.00/0.03	Basic Processor:
0.00/0.03	-> Pairs:
0.00/0.03	 Empty
0.00/0.03	-> Rules:
0.00/0.03	 from(X) -> cons(X,from(s(X)))
0.00/0.03	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.03	 minus(X,0) -> 0
0.00/0.03	 quot(0,s(Y)) -> 0
0.00/0.03	 quot(s(X),s(Y)) -> s(quot(minus(X,Y),s(Y)))
0.00/0.03	 sel(0,cons(X,XS)) -> X
0.00/0.03	 sel(s(N),cons(X,XS)) -> sel(N,XS)
0.00/0.03	 zWquot(cons(X,XS),cons(Y,YS)) -> cons(quot(X,Y),zWquot(XS,YS))
0.00/0.03	 zWquot(nil,XS) -> nil
0.00/0.03	 zWquot(XS,nil) -> nil
0.00/0.03	-> Unhiding rules:
0.00/0.03	 Empty
0.00/0.03	-> Result:
0.00/0.03	 Set P is empty
0.00/0.03	
0.00/0.03	The problem is finite.
0.00/0.04	EOF