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)
0.00/0.01	(and 1)
0.00/0.01	(first 1 2)
0.00/0.01	(from)
0.00/0.01	(if 1)
0.00/0.01	(0)
0.00/0.01	(cons)
0.00/0.01	(false)
0.00/0.01	(nil)
0.00/0.01	(s)
0.00/0.01	(true)
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	and(false,Y) -> false
0.00/0.01	and(true,X) -> X
0.00/0.01	first(0,X) -> nil
0.00/0.01	first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
0.00/0.01	from(X) -> cons(X,from(s(X)))
0.00/0.01	if(false,X,Y) -> Y
0.00/0.01	if(true,X,Y) -> X
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	 and(false,Y) -> false
0.00/0.01	 and(true,X) -> X
0.00/0.01	 first(0,X) -> nil
0.00/0.01	 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 if(false,X,Y) -> Y
0.00/0.01	 if(true,X,Y) -> X
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	 ADD(0,X) -> X
0.00/0.01	 AND(true,X) -> X
0.00/0.01	 IF(false,X,Y) -> Y
0.00/0.01	 IF(true,X,Y) -> X
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	 and(false,Y) -> false
0.00/0.01	 and(true,X) -> X
0.00/0.01	 first(0,X) -> nil
0.00/0.01	 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 if(false,X,Y) -> Y
0.00/0.01	 if(true,X,Y) -> X
0.00/0.01	-> Unhiding Rules:
0.00/0.01	 add(X,Y) -> ADD(X,Y)
0.00/0.01	 add(x3,Y) -> x3
0.00/0.01	 first(X,Z) -> FIRST(X,Z)
0.00/0.01	 first(X,x3) -> x3
0.00/0.01	 first(x3,Z) -> x3
0.00/0.01	 from(s(X)) -> FROM(s(X))
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	 ADD(0,X) -> X
0.00/0.01	 AND(true,X) -> X
0.00/0.01	 IF(false,X,Y) -> Y
0.00/0.01	 IF(true,X,Y) -> X
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	 and(false,Y) -> false
0.00/0.01	 and(true,X) -> X
0.00/0.01	 first(0,X) -> nil
0.00/0.01	 first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
0.00/0.01	 from(X) -> cons(X,from(s(X)))
0.00/0.01	 if(false,X,Y) -> Y
0.00/0.01	 if(true,X,Y) -> X
0.00/0.01	-> Unhiding rules:
0.00/0.01	 add(X,Y) -> ADD(X,Y)
0.00/0.01	 add(x3,Y) -> x3
0.00/0.01	 first(X,Z) -> FIRST(X,Z)
0.00/0.01	 first(X,x3) -> x3
0.00/0.01	 first(x3,Z) -> x3
0.00/0.01	 from(s(X)) -> FROM(s(X))
0.00/0.01	-> Result:
0.00/0.01	 All pairs P are from Px1
0.00/0.01	
0.00/0.01	The problem is finite.
0.00/0.01	EOF