0.00/0.08	YES
0.00/0.08	We consider the system theBenchmark.
0.00/0.08	
0.00/0.08	  Alphabet:
0.00/0.08	
0.00/0.08	    cons : [a * c] --> c 
0.00/0.08	    dropWhile : [a -> b * c] --> c 
0.00/0.08	    if : [b * c * c] --> c 
0.00/0.08	    nil : [] --> c 
0.00/0.08	    takeWhile : [a -> b * c] --> c 
0.00/0.08	    true : [] --> b 
0.00/0.08	
0.00/0.08	  Rules:
0.00/0.08	
0.00/0.08	    if(true, x, y) => x 
0.00/0.08	    if(true, x, y) => y 
0.00/0.08	    takeWhile(f, nil) => nil 
0.00/0.08	    takeWhile(f, cons(x, y)) => if(f x, cons(x, takeWhile(f, y)), nil) 
0.00/0.08	    dropWhile(f, nil) => nil 
0.00/0.08	    dropWhile(f, cons(x, y)) => if(f x, dropWhile(f, y), cons(x, y)) 
0.00/0.08	
0.00/0.08	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.08	
0.00/0.08	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.08	
0.00/0.08	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.08	
0.00/0.08	  if(true, X, Y) >? X 
0.00/0.08	  if(true, X, Y) >? Y 
0.00/0.08	  takeWhile(F, nil) >? nil 
0.00/0.08	  takeWhile(F, cons(X, Y)) >? if(F X, cons(X, takeWhile(F, Y)), nil) 
0.00/0.08	  dropWhile(F, nil) >? nil 
0.00/0.08	  dropWhile(F, cons(X, Y)) >? if(F X, dropWhile(F, Y), cons(X, Y)) 
0.00/0.08	
0.00/0.08	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.08	
0.00/0.08	The following interpretation satisfies the requirements:
0.00/0.08	
0.00/0.08	  cons = \y0y1.2 + y0 + 2y1 
0.00/0.08	  dropWhile = \G0y1.3 + 2y1 + 2y1G0(y1) + 2G0(0) 
0.00/0.08	  if = \y0y1y2.y0 + y1 + y2 
0.00/0.08	  nil = 0 
0.00/0.08	  takeWhile = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
0.00/0.08	  true = 3 
0.00/0.08	
0.00/0.08	Using this interpretation, the requirements translate to:
0.00/0.08	
0.00/0.08	  [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x0 = [[_x0]] 
0.00/0.08	  [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x1 = [[_x1]] 
0.00/0.08	  [[takeWhile(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
0.00/0.08	  [[takeWhile(_F0, cons(_x1, _x2))]] = 6 + 3x1 + 6x2 + F0(0) + 2x1F0(2 + x1 + 2x2) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 2 + 2x1 + 6x2 + F0(x1) + 2F0(0) + 4x2F0(x2) = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] 
0.00/0.08	  [[dropWhile(_F0, nil)]] = 3 + 2F0(0) > 0 = [[nil]] 
0.00/0.08	  [[dropWhile(_F0, cons(_x1, _x2))]] = 7 + 2x1 + 4x2 + 2x1F0(2 + x1 + 2x2) + 2F0(0) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 5 + 2x1 + 4x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _x2))]] 
0.00/0.08	
0.00/0.08	We can thus remove the following rules:
0.00/0.08	
0.00/0.08	  if(true, X, Y) => X 
0.00/0.08	  if(true, X, Y) => Y 
0.00/0.08	  takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) 
0.00/0.08	  dropWhile(F, nil) => nil 
0.00/0.08	  dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(X, Y)) 
0.00/0.08	
0.00/0.08	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.08	
0.00/0.08	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.08	
0.00/0.08	  takeWhile(F, nil) >? nil 
0.00/0.08	
0.00/0.08	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.08	
0.00/0.08	The following interpretation satisfies the requirements:
0.00/0.08	
0.00/0.08	  nil = 0 
0.00/0.08	  takeWhile = \G0y1.3 + 3y1 + G0(0) 
0.00/0.08	
0.00/0.08	Using this interpretation, the requirements translate to:
0.00/0.08	
0.00/0.08	  [[takeWhile(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.08	
0.00/0.08	We can thus remove the following rules:
0.00/0.08	
0.00/0.08	  takeWhile(F, nil) => nil 
0.00/0.08	
0.00/0.08	All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.
0.00/0.08	
0.00/0.08	
0.00/0.08	+++ Citations +++
0.00/0.08	
0.00/0.08	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.08	EOF