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 : [] --> b 
0.00/0.04	    cons : [a * b] --> b 
0.00/0.04	    nil : [] --> b 
0.00/0.04	    plus : [b * b] --> b 
0.00/0.04	    s : [b] --> b 
0.00/0.04	    sumwith : [a -> b * b] --> b 
0.00/0.04	
0.00/0.04	  Rules:
0.00/0.04	
0.00/0.04	    plus(0, x) => x 
0.00/0.04	    plus(s(x), y) => s(plus(x, y)) 
0.00/0.04	    sumwith(f, nil) => nil 
0.00/0.04	    sumwith(f, cons(x, y)) => plus(f x, sumwith(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	  plus(0, X) >? X 
0.00/0.04	  plus(s(X), Y) >? s(plus(X, Y)) 
0.00/0.04	  sumwith(F, nil) >? nil 
0.00/0.04	  sumwith(F, cons(X, Y)) >? plus(F X, sumwith(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	  cons = \y0y1.3 + 3y0 + 3y1 
0.00/0.04	  nil = 0 
0.00/0.04	  plus = \y0y1.1 + y0 + y1 
0.00/0.04	  s = \y0.y0 
0.00/0.04	  sumwith = \G0y1.3 + 3y1 + G0(0) + 2y1G0(y1) 
0.00/0.04	
0.00/0.04	Using this interpretation, the requirements translate to:
0.00/0.04	
0.00/0.04	  [[plus(0, _x0)]] = 4 + x0 > x0 = [[_x0]] 
0.00/0.04	  [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 
0.00/0.04	  [[sumwith(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.04	  [[sumwith(_F0, cons(_x1, _x2))]] = 12 + 9x1 + 9x2 + F0(0) + 6x1F0(3 + 3x1 + 3x2) + 6x2F0(3 + 3x1 + 3x2) + 6F0(3 + 3x1 + 3x2) > 4 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[plus(_F0 _x1, sumwith(_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	  plus(0, X) => X 
0.00/0.04	  sumwith(F, nil) => nil 
0.00/0.04	  sumwith(F, cons(X, Y)) => plus(F X, sumwith(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	  plus(s(X), Y) >? s(plus(X, 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	  plus = \y0y1.y1 + 3y0 
0.00/0.04	  s = \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	  [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 > 1 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
0.00/0.04	
0.00/0.04	We can thus remove the following rules:
0.00/0.04	
0.00/0.04	  plus(s(X), Y) => s(plus(X, 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