0.00/0.07	YES
0.00/0.07	We consider the system theBenchmark.
0.00/0.07	
0.00/0.07	  Alphabet:
0.00/0.07	
0.00/0.07	    0 : [] --> o 
0.00/0.07	    either : [o * o] --> o 
0.00/0.07	    f : [o -> o * o * o] --> o 
0.00/0.07	    g : [o * o] --> o 
0.00/0.07	    s : [o] --> o 
0.00/0.07	
0.00/0.07	  Rules:
0.00/0.07	
0.00/0.07	    f(h, x, 0) => 0 
0.00/0.07	    f(h, x, s(y)) => g(y, either(y, h x)) 
0.00/0.07	    g(x, y) => f(/\z.s(0), y, x) 
0.00/0.07	
0.00/0.07	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.07	
0.00/0.07	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.07	
0.00/0.07	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.07	
0.00/0.07	  f(F, X, 0) >? 0 
0.00/0.07	  f(F, X, s(Y)) >? g(Y, either(Y, F X)) 
0.00/0.07	  g(X, Y) >? f(/\x.s(0), Y, X) 
0.00/0.07	
0.00/0.07	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.07	
0.00/0.07	The following interpretation satisfies the requirements:
0.00/0.07	
0.00/0.07	  0 = 0 
0.00/0.07	  either = \y0y1.y0 + y1 
0.00/0.07	  f = \G0y1y2.1 + y1 + 2y2 + G0(y1) 
0.00/0.07	  g = \y0y1.1 + y1 + 2y0 
0.00/0.07	  s = \y0.2y0 
0.00/0.07	
0.00/0.07	Using this interpretation, the requirements translate to:
0.00/0.07	
0.00/0.07	  [[f(_F0, _x1, 0)]] = 1 + x1 + F0(x1) > 0 = [[0]] 
0.00/0.07	  [[f(_F0, _x1, s(_x2))]] = 1 + x1 + 4x2 + F0(x1) >= 1 + x1 + 3x2 + F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] 
0.00/0.08	  [[g(_x0, _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[f(/\x.s(0), _x1, _x0)]] 
0.00/0.08	
0.00/0.08	We can thus remove the following rules:
0.00/0.08	
0.00/0.08	  f(F, X, 0) => 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	  f(F, X, s(Y)) >? g(Y, either(Y, F X)) 
0.00/0.08	  g(X, Y) >? f(/\x.s(0), Y, X) 
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	  0 = 0 
0.00/0.08	  either = \y0y1.y0 + y1 
0.00/0.08	  f = \G0y1y2.2y1 + 3y2 + 2G0(y1) 
0.00/0.08	  g = \y0y1.2 + 2y1 + 3y0 
0.00/0.08	  s = \y0.1 + 3y0 
0.00/0.08	
0.00/0.08	Using this interpretation, the requirements translate to:
0.00/0.08	
0.00/0.08	  [[f(_F0, _x1, s(_x2))]] = 3 + 2x1 + 9x2 + 2F0(x1) > 2 + 2x1 + 5x2 + 2F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] 
0.00/0.08	  [[g(_x0, _x1)]] = 2 + 2x1 + 3x0 >= 2 + 2x1 + 3x0 = [[f(/\x.s(0), _x1, _x0)]] 
0.00/0.08	
0.00/0.08	We can thus remove the following rules:
0.00/0.08	
0.00/0.08	  f(F, X, s(Y)) => g(Y, either(Y, F X)) 
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	  g(X, Y) >? f(/\x.s(0), Y, X) 
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	  0 = 0 
0.00/0.08	  f = \G0y1y2.y1 + y2 + G0(0) 
0.00/0.08	  g = \y0y1.3 + 3y0 + 3y1 
0.00/0.08	  s = \y0.y0 
0.00/0.08	
0.00/0.08	Using this interpretation, the requirements translate to:
0.00/0.08	
0.00/0.08	  [[g(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f(/\x.s(0), _x1, _x0)]] 
0.00/0.08	
0.00/0.08	We can thus remove the following rules:
0.00/0.08	
0.00/0.08	  g(X, Y) => f(/\x.s(0), Y, X) 
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