0.00/0.03	YES
0.00/0.03	We consider the system theBenchmark.
0.00/0.03	
0.00/0.03	  Alphabet:
0.00/0.03	
0.00/0.03	    cons : [a * list] --> list 
0.00/0.03	    map : [list * a -> a] --> list 
0.00/0.03	    nil : [] --> list 
0.00/0.03	
0.00/0.03	  Rules:
0.00/0.03	
0.00/0.03	    map(nil, f) => nil 
0.00/0.03	    map(cons(x, y), f) => cons(f x, map(y, f)) 
0.00/0.03	
0.00/0.03	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.03	
0.00/0.03	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.03	
0.00/0.03	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.03	
0.00/0.03	  map(nil, F) >? nil 
0.00/0.03	  map(cons(X, Y), F) >? cons(F X, map(Y, F)) 
0.00/0.03	
0.00/0.03	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.03	
0.00/0.03	The following interpretation satisfies the requirements:
0.00/0.03	
0.00/0.03	  cons = \y0y1.3 + y0 + y1 
0.00/0.03	  map = \y0G1.3 + 3y0 + G1(y0) + 3y0G1(y0) 
0.00/0.03	  nil = 2 
0.00/0.03	
0.00/0.03	Using this interpretation, the requirements translate to:
0.00/0.03	
0.00/0.03	  [[map(nil, _F0)]] = 9 + 7F0(2) > 2 = [[nil]] 
0.00/0.03	  [[map(cons(_x0, _x1), _F2)]] = 12 + 3x0 + 3x1 + 3x0F2(3 + x0 + x1) + 3x1F2(3 + x0 + x1) + 10F2(3 + x0 + x1) > 6 + x0 + 3x1 + F2(x0) + F2(x1) + 3x1F2(x1) = [[cons(_F2 _x0, map(_x1, _F2))]] 
0.00/0.03	
0.00/0.03	We can thus remove the following rules:
0.00/0.03	
0.00/0.03	  map(nil, F) => nil 
0.00/0.03	  map(cons(X, Y), F) => cons(F X, map(Y, F)) 
0.00/0.03	
0.00/0.03	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.
0.00/0.03	
0.00/0.03	
0.00/0.03	+++ Citations +++
0.00/0.03	
0.00/0.03	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.03	EOF