0.00/0.01	YES
0.00/0.01	We consider the system theBenchmark.
0.00/0.01	
0.00/0.01	  Alphabet:
0.00/0.01	
0.00/0.01	    minus : [N * N] --> N 
0.00/0.01	    s : [N] --> N 
0.00/0.01	    z : [] --> N 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    minus(z, x) => z 
0.00/0.01	    minus(x, z) => x 
0.00/0.01	    minus(s(x), s(y)) => minus(x, y) 
0.00/0.01	    minus(x, x) => z 
0.00/0.01	
0.00/0.01	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.01	
0.00/0.01	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.01	
0.00/0.01	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.01	
0.00/0.01	  minus(z, X) >? z 
0.00/0.01	  minus(X, z) >? X 
0.00/0.01	  minus(s(X), s(Y)) >? minus(X, Y) 
0.00/0.01	  minus(X, X) >? z 
0.00/0.01	
0.00/0.01	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.01	
0.00/0.01	The following interpretation satisfies the requirements:
0.00/0.01	
0.00/0.01	  minus = \y0y1.3 + y1 + 3y0 
0.00/0.01	  s = \y0.3 + 3y0 
0.00/0.01	  z = 0 
0.00/0.01	
0.00/0.01	Using this interpretation, the requirements translate to:
0.00/0.01	
0.00/0.01	  [[minus(z, _x0)]] = 3 + x0 > 0 = [[z]] 
0.00/0.01	  [[minus(_x0, z)]] = 3 + 3x0 > x0 = [[_x0]] 
0.00/0.01	  [[minus(s(_x0), s(_x1))]] = 15 + 3x1 + 9x0 > 3 + x1 + 3x0 = [[minus(_x0, _x1)]] 
0.00/0.01	  [[minus(_x0, _x0)]] = 3 + 4x0 > 0 = [[z]] 
0.00/0.01	
0.00/0.01	We can thus remove the following rules:
0.00/0.01	
0.00/0.01	  minus(z, X) => z 
0.00/0.01	  minus(X, z) => X 
0.00/0.01	  minus(s(X), s(Y)) => minus(X, Y) 
0.00/0.01	  minus(X, X) => z 
0.00/0.01	
0.00/0.01	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.
0.00/0.01	
0.00/0.01	
0.00/0.01	+++ Citations +++
0.00/0.01	
0.00/0.01	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.01	EOF