0.00/0.04	YES
0.00/0.04	We consider the system theBenchmark.
0.00/0.04	
0.00/0.04	  Alphabet:
0.00/0.04	
0.00/0.04	    cons : [a * c] --> c 
0.00/0.04	    consif : [b * a * c] --> c 
0.00/0.04	    false : [] --> b 
0.00/0.04	    filter : [a -> b * c] --> c 
0.00/0.04	    nil : [] --> c 
0.00/0.04	    true : [] --> b 
0.00/0.04	
0.00/0.04	  Rules:
0.00/0.04	
0.00/0.04	    consif(true, x, y) => cons(x, y) 
0.00/0.04	    consif(false, x, y) => y 
0.00/0.04	    filter(f, nil) => nil 
0.00/0.04	    filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) 
0.00/0.04	
0.00/0.04	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.04	
0.00/0.04	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.04	
0.00/0.04	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.04	
0.00/0.04	  consif(true, X, Y) >? cons(X, Y) 
0.00/0.04	  consif(false, X, Y) >? Y 
0.00/0.04	  filter(F, nil) >? nil 
0.00/0.04	  filter(F, cons(X, Y)) >? consif(F X, X, filter(F, Y)) 
0.00/0.04	
0.00/0.04	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.04	
0.00/0.04	The following interpretation satisfies the requirements:
0.00/0.04	
0.00/0.04	  cons = \y0y1.2 + y0 + y1 
0.00/0.04	  consif = \y0y1y2.y0 + y1 + y2 
0.00/0.04	  false = 3 
0.00/0.04	  filter = \G0y1.2y1 + G0(y1) + y1G0(y1) 
0.00/0.04	  nil = 1 
0.00/0.04	  true = 3 
0.00/0.04	
0.00/0.04	Using this interpretation, the requirements translate to:
0.00/0.04	
0.00/0.04	  [[consif(true, _x0, _x1)]] = 3 + x0 + x1 > 2 + x0 + x1 = [[cons(_x0, _x1)]] 
0.00/0.04	  [[consif(false, _x0, _x1)]] = 3 + x0 + x1 > x1 = [[_x1]] 
0.00/0.04	  [[filter(_F0, nil)]] = 2 + 2F0(1) > 1 = [[nil]] 
0.00/0.04	  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 3F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > 2x1 + 2x2 + F0(x1) + F0(x2) + x2F0(x2) = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] 
0.00/0.04	
0.00/0.04	We can thus remove the following rules:
0.00/0.04	
0.00/0.04	  consif(true, X, Y) => cons(X, Y) 
0.00/0.04	  consif(false, X, Y) => Y 
0.00/0.04	  filter(F, nil) => nil 
0.00/0.04	  filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 
0.00/0.04	
0.00/0.04	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.
0.00/0.04	
0.00/0.04	
0.00/0.04	+++ Citations +++
0.00/0.04	
0.00/0.04	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.04	EOF