0.00/0.04	YES
0.00/0.04	
0.00/0.04	Problem 1: 
0.00/0.04	
0.00/0.04	(VAR X Y)
0.00/0.04	(STRATEGY CONTEXTSENSITIVE
0.00/0.04	(div 1)
0.00/0.04	(geq)
0.00/0.04	(if 1)
0.00/0.04	(minus)
0.00/0.04	(0)
0.00/0.04	(false)
0.00/0.04	(s 1)
0.00/0.04	(true)
0.00/0.04	)
0.00/0.04	(RULES
0.00/0.04	div(0,s(Y)) -> 0
0.00/0.04	div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	geq(0,s(Y)) -> false
0.00/0.04	geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	geq(X,0) -> true
0.00/0.04	if(false,X,Y) -> Y
0.00/0.04	if(true,X,Y) -> X
0.00/0.04	minus(0,Y) -> 0
0.00/0.04	minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	)
0.00/0.04	
0.00/0.04	Problem 1: 
0.00/0.04	
0.00/0.04	Innermost Equivalent Processor:
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> The context-sensitive term rewriting system is an orthogonal system. Therefore, innermost cs-termination implies cs-termination.
0.00/0.04	 
0.00/0.04	
0.00/0.04	Problem 1: 
0.00/0.04	
0.00/0.04	Dependency Pairs Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 DIV(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	 DIV(s(X),s(Y)) -> IF(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 GEQ(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	 IF(false,X,Y) -> Y
0.00/0.04	 IF(true,X,Y) -> X
0.00/0.04	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding Rules:
0.00/0.04	 div(minus(X,Y),s(Y)) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 div(minus(X,Y),s(Y)) -> MINUS(X,Y)
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> MINUS(X,Y)
0.00/0.04	
0.00/0.04	Problem 1: 
0.00/0.04	
0.00/0.04	SCC Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 DIV(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	 DIV(s(X),s(Y)) -> IF(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 GEQ(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	 IF(false,X,Y) -> Y
0.00/0.04	 IF(true,X,Y) -> X
0.00/0.04	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 div(minus(X,Y),s(Y)) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 div(minus(X,Y),s(Y)) -> MINUS(X,Y)
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> MINUS(X,Y)
0.00/0.04	->Strongly Connected Components:
0.00/0.04	->->Cycle:
0.00/0.04	->->-> Pairs:
0.00/0.04	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.04	->->-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	->->-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	->->Cycle:
0.00/0.04	->->-> Pairs:
0.00/0.04	 GEQ(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	->->-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	->->-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	->->Cycle:
0.00/0.04	->->-> Pairs:
0.00/0.04	 DIV(s(X),s(Y)) -> IF(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 IF(false,X,Y) -> Y
0.00/0.04	 IF(true,X,Y) -> X
0.00/0.04	->->-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	->->-> Unhiding rules:
0.00/0.04	 div(minus(X,Y),s(Y)) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> DIV(minus(X,Y),s(Y))
0.00/0.04	
0.00/0.04	
0.00/0.04	The problem is decomposed in 3 subproblems.
0.00/0.04	
0.00/0.04	Problem 1.1: 
0.00/0.04	
0.00/0.04	SubNColl Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 MINUS(s(X),s(Y)) -> MINUS(X,Y)
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	->Projection:
0.00/0.04	 pi(MINUS) = 1
0.00/0.04	
0.00/0.04	Problem 1.1: 
0.00/0.04	
0.00/0.04	Basic Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 Empty
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	-> Result:
0.00/0.04	 Set P is empty
0.00/0.04	
0.00/0.04	The problem is finite.
0.00/0.04	
0.00/0.04	Problem 1.2: 
0.00/0.04	
0.00/0.04	SubNColl Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 GEQ(s(X),s(Y)) -> GEQ(X,Y)
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	->Projection:
0.00/0.04	 pi(GEQ) = 1
0.00/0.04	
0.00/0.04	Problem 1.2: 
0.00/0.04	
0.00/0.04	Basic Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 Empty
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 Empty
0.00/0.04	-> Result:
0.00/0.04	 Set P is empty
0.00/0.04	
0.00/0.04	The problem is finite.
0.00/0.04	
0.00/0.04	Problem 1.3: 
0.00/0.04	
0.00/0.04	Reduction Pairs Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 DIV(s(X),s(Y)) -> IF(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 IF(false,X,Y) -> Y
0.00/0.04	 IF(true,X,Y) -> X
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 div(minus(X,Y),s(Y)) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> DIV(minus(X,Y),s(Y))
0.00/0.04	-> Usable rules:
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	->Interpretation type:
0.00/0.04	 Linear
0.00/0.04	->Coefficients:
0.00/0.04	 Natural Numbers
0.00/0.04	->Dimension:
0.00/0.04	 1
0.00/0.04	->Bound:
0.00/0.04	 2
0.00/0.04	->Interpretation:
0.00/0.04	 
0.00/0.04	[div](X1,X2) = 2.X2 + 2
0.00/0.04	[geq](X1,X2) = 2
0.00/0.04	[minus](X1,X2) = 0
0.00/0.04	[0] = 0
0.00/0.04	[false] = 2
0.00/0.04	[s](X) = X + 2
0.00/0.04	[true] = 2
0.00/0.04	[DIV](X1,X2) = 2.X1 + 2.X2 + 2
0.00/0.04	[IF](X1,X2,X3) = X2 + 2.X3 + 1
0.00/0.04	
0.00/0.04	Problem 1.3: 
0.00/0.04	
0.00/0.04	Basic Processor:
0.00/0.04	-> Pairs:
0.00/0.04	 IF(false,X,Y) -> Y
0.00/0.04	 IF(true,X,Y) -> X
0.00/0.04	-> Rules:
0.00/0.04	 div(0,s(Y)) -> 0
0.00/0.04	 div(s(X),s(Y)) -> if(geq(X,Y),s(div(minus(X,Y),s(Y))),0)
0.00/0.04	 geq(0,s(Y)) -> false
0.00/0.04	 geq(s(X),s(Y)) -> geq(X,Y)
0.00/0.04	 geq(X,0) -> true
0.00/0.04	 if(false,X,Y) -> Y
0.00/0.04	 if(true,X,Y) -> X
0.00/0.04	 minus(0,Y) -> 0
0.00/0.04	 minus(s(X),s(Y)) -> minus(X,Y)
0.00/0.04	-> Unhiding rules:
0.00/0.04	 div(minus(X,Y),s(Y)) -> DIV(minus(X,Y),s(Y))
0.00/0.04	 s(div(minus(X,Y),s(Y))) -> DIV(minus(X,Y),s(Y))
0.00/0.04	-> Result:
0.00/0.04	 All pairs P are from Px1
0.00/0.04	
0.00/0.04	The problem is finite.
0.00/0.05	EOF