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	    and : [c * c] --> c 
0.00/0.01	    arrow : [t * t] --> t 
0.00/0.01	    lessthan : [t * t] --> c 
0.00/0.01	
0.00/0.01	  Rules:
0.00/0.01	
0.00/0.01	    lessthan(arrow(x, y), arrow(z, u)) => and(lessthan(z, x), lessthan(y, u)) 
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	  lessthan(arrow(X, Y), arrow(Z, U)) >? and(lessthan(Z, X), lessthan(Y, U)) 
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	  and = \y0y1.y0 + y1 
0.00/0.01	  arrow = \y0y1.3 + 3y0 + 3y1 
0.00/0.01	  lessthan = \y0y1.2y0 + 3y1 
0.00/0.01	
0.00/0.01	Using this interpretation, the requirements translate to:
0.00/0.01	
0.00/0.01	  [[lessthan(arrow(_x0, _x1), arrow(_x2, _x3))]] = 15 + 6x0 + 6x1 + 9x2 + 9x3 > 2x1 + 2x2 + 3x0 + 3x3 = [[and(lessthan(_x2, _x0), lessthan(_x1, _x3))]] 
0.00/0.01	
0.00/0.01	We can thus remove the following rules:
0.00/0.01	
0.00/0.01	  lessthan(arrow(X, Y), arrow(Z, U)) => and(lessthan(Z, X), lessthan(Y, U)) 
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