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	    0 : [] --> a 
0.00/0.04	    add : [a] --> a -> a 
0.00/0.04	    cons : [b * c] --> c 
0.00/0.04	    id : [] --> a -> a 
0.00/0.04	    map : [b -> b * c] --> c 
0.00/0.04	    nil : [] --> c 
0.00/0.04	    s : [a] --> a 
0.00/0.04	
0.00/0.04	  Rules:
0.00/0.04	
0.00/0.04	    id x => x 
0.00/0.04	    add(0) => id 
0.00/0.04	    add(s(x)) y => s(add(x) y) 
0.00/0.04	    map(f, nil) => nil 
0.00/0.04	    map(f, cons(x, y)) => cons(f x, map(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	  id X >? X 
0.00/0.04	  add(0) >? id 
0.00/0.04	  add(s(X)) Y >? s(add(X) Y) 
0.00/0.04	  map(F, nil) >? nil 
0.00/0.04	  map(F, cons(X, Y)) >? cons(F X, map(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	  0 = 3 
0.00/0.04	  add = \y0y1.3 + 3y0 + 3y1 
0.00/0.04	  cons = \y0y1.3 + y0 + y1 
0.00/0.04	  id = \y0.3y0 
0.00/0.04	  map = \G0y1.3 + 3y1 + G0(y1) + 3y1G0(y1) 
0.00/0.04	  nil = 0 
0.00/0.04	  s = \y0.3 + y0 
0.00/0.04	
0.00/0.04	Using this interpretation, the requirements translate to:
0.00/0.04	
0.00/0.04	  [[id _x0]] = 4x0 >= x0 = [[_x0]] 
0.00/0.04	  [[add(0)]] = \y0.12 + 3y0 > \y0.3y0 = [[id]] 
0.00/0.04	  [[add(s(_x0)) _x1]] = 12 + 3x0 + 4x1 > 6 + 3x0 + 4x1 = [[s(add(_x0) _x1)]] 
0.00/0.04	  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.04	  [[map(_F0, cons(_x1, _x2))]] = 12 + 3x1 + 3x2 + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 10F0(3 + x1 + x2) > 6 + x1 + 3x2 + F0(x1) + F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_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	  add(0) => id 
0.00/0.04	  add(s(X)) Y => s(add(X) Y) 
0.00/0.04	  map(F, nil) => nil 
0.00/0.04	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
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	  id(X) >? X 
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	  id = \y0.1 + y0 
0.00/0.04	
0.00/0.04	Using this interpretation, the requirements translate to:
0.00/0.04	
0.00/0.04	  [[id(_x0)]] = 1 + x0 > x0 = [[_x0]] 
0.00/0.04	
0.00/0.04	We can thus remove the following rules:
0.00/0.04	
0.00/0.04	  id(X) => X 
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