0.00/0.08	YES
0.00/0.09	We consider the system theBenchmark.
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    cons : [b * c] --> c 
0.00/0.09	    leaf : [a] --> b 
0.00/0.09	    mapt : [a -> a * b] --> b 
0.00/0.09	    maptlist : [a -> a * c] --> c 
0.00/0.09	    nil : [] --> c 
0.00/0.09	    node : [c] --> b 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    mapt(f, leaf(x)) => leaf(f x) 
0.00/0.09	    mapt(f, node(x)) => node(maptlist(f, x)) 
0.00/0.09	    maptlist(f, nil) => nil 
0.00/0.09	    maptlist(f, cons(x, y)) => cons(mapt(f, x), maptlist(f, y)) 
0.00/0.09	
0.00/0.09	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.09	
0.00/0.09	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.09	
0.00/0.09	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.09	
0.00/0.09	  mapt(F, leaf(X)) >? leaf(F X) 
0.00/0.09	  mapt(F, node(X)) >? node(maptlist(F, X)) 
0.00/0.09	  maptlist(F, nil) >? nil 
0.00/0.09	  maptlist(F, cons(X, Y)) >? cons(mapt(F, X), maptlist(F, Y)) 
0.00/0.09	
0.00/0.09	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.09	
0.00/0.09	The following interpretation satisfies the requirements:
0.00/0.09	
0.00/0.09	  cons = \y0y1.2 + y1 + 2y0 
0.00/0.09	  leaf = \y0.y0 
0.00/0.09	  mapt = \G0y1.1 + 3y1 + G0(y1) + 2y1G0(y1) 
0.00/0.09	  maptlist = \G0y1.3y1 + G0(0) + G0(y1) + 2y1G0(y1) 
0.00/0.09	  nil = 0 
0.00/0.09	  node = \y0.1 + y0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[mapt(_F0, leaf(_x1))]] = 1 + 3x1 + F0(x1) + 2x1F0(x1) > x1 + F0(x1) = [[leaf(_F0 _x1)]] 
0.00/0.09	  [[mapt(_F0, node(_x1))]] = 4 + 3x1 + 2x1F0(1 + x1) + 3F0(1 + x1) > 1 + 3x1 + F0(0) + F0(x1) + 2x1F0(x1) = [[node(maptlist(_F0, _x1))]] 
0.00/0.09	  [[maptlist(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
0.00/0.09	  [[maptlist(_F0, cons(_x1, _x2))]] = 6 + 3x2 + 6x1 + F0(0) + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 5F0(2 + x2 + 2x1) > 4 + 3x2 + 6x1 + F0(0) + F0(x2) + 2x2F0(x2) + 2F0(x1) + 4x1F0(x1) = [[cons(mapt(_F0, _x1), maptlist(_F0, _x2))]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  mapt(F, leaf(X)) => leaf(F X) 
0.00/0.09	  mapt(F, node(X)) => node(maptlist(F, X)) 
0.00/0.09	  maptlist(F, cons(X, Y)) => cons(mapt(F, X), maptlist(F, Y)) 
0.00/0.09	
0.00/0.09	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.09	
0.00/0.09	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.09	
0.00/0.09	  maptlist(F, nil) >? nil 
0.00/0.09	
0.00/0.09	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.09	
0.00/0.09	The following interpretation satisfies the requirements:
0.00/0.09	
0.00/0.09	  maptlist = \G0y1.3 + 3y1 + G0(0) 
0.00/0.09	  nil = 0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[maptlist(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  maptlist(F, nil) => nil 
0.00/0.09	
0.00/0.09	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.
0.00/0.09	
0.00/0.09	
0.00/0.09	+++ Citations +++
0.00/0.09	
0.00/0.09	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.09	EOF