0.00/0.01	YES
0.00/0.01	
0.00/0.01	Problem 1: 
0.00/0.01	
0.00/0.01	(VAR X Y Z)
0.00/0.01	(STRATEGY CONTEXTSENSITIVE
0.00/0.01	(add 1 2)
0.00/0.01	(from 1)
0.00/0.01	(fst 1 2)
0.00/0.01	(len 1)
0.00/0.01	(0)
0.00/0.01	(cons 1)
0.00/0.01	(nil)
0.00/0.01	(s)
0.00/0.01	)
0.00/0.01	(RULES
0.00/0.01	add(0,X) -> X
0.00/0.01	add(s(X),Y) -> s(add(X,Y))
0.00/0.01	from(X) -> cons(X,from(s(X)))
0.00/0.01	fst(0,Z) -> nil
0.00/0.01	fst(s(X),cons(Y,Z)) -> cons(Y,fst(X,Z))
0.00/0.01	len(cons(X,Z)) -> s(len(Z))
0.00/0.01	len(nil) -> 0
0.00/0.01	)
0.00/0.01	
0.00/0.01	Problem 1: 
0.00/0.01	
0.00/0.01	Innermost Equivalent Processor:
0.00/0.01	-> Rules:
0.00/0.01	 add(0,X) -> X
0.00/0.01	 add(s(X),Y) -> s(add(X,Y))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 fst(0,Z) -> nil
0.00/0.01	 fst(s(X),cons(Y,Z)) -> cons(Y,fst(X,Z))
0.00/0.01	 len(cons(X,Z)) -> s(len(Z))
0.00/0.01	 len(nil) -> 0
0.00/0.01	-> The context-sensitive term rewriting system is an orthogonal system. Therefore, innermost cs-termination implies cs-termination.
0.00/0.01	 
0.00/0.01	
0.00/0.01	Problem 1: 
0.00/0.01	
0.00/0.01	Dependency Pairs Processor:
0.00/0.01	-> Pairs:
0.00/0.01	 Empty
0.00/0.01	-> Rules:
0.00/0.01	 add(0,X) -> X
0.00/0.01	 add(s(X),Y) -> s(add(X,Y))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 fst(0,Z) -> nil
0.00/0.01	 fst(s(X),cons(Y,Z)) -> cons(Y,fst(X,Z))
0.00/0.01	 len(cons(X,Z)) -> s(len(Z))
0.00/0.01	 len(nil) -> 0
0.00/0.01	-> Unhiding Rules:
0.00/0.01	 Empty
0.00/0.01	
0.00/0.01	Problem 1: 
0.00/0.01	
0.00/0.01	Basic Processor:
0.00/0.01	-> Pairs:
0.00/0.01	 Empty
0.00/0.01	-> Rules:
0.00/0.01	 add(0,X) -> X
0.00/0.01	 add(s(X),Y) -> s(add(X,Y))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 fst(0,Z) -> nil
0.00/0.01	 fst(s(X),cons(Y,Z)) -> cons(Y,fst(X,Z))
0.00/0.01	 len(cons(X,Z)) -> s(len(Z))
0.00/0.01	 len(nil) -> 0
0.00/0.01	-> Unhiding rules:
0.00/0.01	 Empty
0.00/0.01	-> Result:
0.00/0.01	 Set P is empty
0.00/0.01	
0.00/0.01	The problem is finite.
0.00/0.01	EOF