0.00/0.08	YES
0.00/0.08	We consider the system theBenchmark.
0.00/0.08	
0.00/0.08	  Alphabet:
0.00/0.08	
0.00/0.08	    0 : [] --> b 
0.00/0.08	    cons : [b * a] --> a 
0.00/0.08	    curry : [b -> b -> b * b] --> b -> b 
0.00/0.08	    inc : [] --> a -> a 
0.00/0.08	    map : [b -> b] --> a -> a 
0.00/0.08	    nil : [] --> a 
0.00/0.08	    plus : [] --> b -> b -> b 
0.00/0.08	    s : [b] --> b 
0.00/0.08	
0.00/0.08	  Rules:
0.00/0.08	
0.00/0.08	    plus 0 x => x 
0.00/0.08	    plus s(x) y => s(plus x y) 
0.00/0.08	    map(f) nil => nil 
0.00/0.08	    map(f) cons(x, y) => cons(f x, map(f) y) 
0.00/0.08	    curry(f, x) y => f x y 
0.00/0.08	    inc => map(curry(plus, s(0))) 
0.00/0.08	
0.00/0.08	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.08	
0.00/0.08	Symbol curry is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:
0.00/0.08	
0.00/0.08	  Alphabet:
0.00/0.08	
0.00/0.08	    0 : [] --> b 
0.00/0.08	    cons : [b * a] --> a 
0.00/0.08	    inc : [] --> a -> a 
0.00/0.08	    map : [b -> b] --> a -> a 
0.00/0.08	    nil : [] --> a 
0.00/0.08	    plus : [b] --> b -> b 
0.00/0.08	    s : [b] --> b 
0.00/0.08	
0.00/0.08	  Rules:
0.00/0.08	
0.00/0.08	    plus(0) X => X 
0.00/0.08	    plus(s(X)) Y => s(plus(X) Y) 
0.00/0.08	    map(F) nil => nil 
0.00/0.08	    map(F) cons(X, Y) => cons(F X, map(F) Y) 
0.00/0.08	    inc => map(plus(s(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	  plus(0) X >? X 
0.00/0.08	  plus(s(X)) Y >? s(plus(X) Y) 
0.00/0.08	  map(F) nil >? nil 
0.00/0.08	  map(F) cons(X, Y) >? cons(F X, map(F) Y) 
0.00/0.08	  inc >? map(plus(s(0))) 
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	  cons = \y0y1.3 + y0 + y1 
0.00/0.08	  inc = \y0.3 + 3y0 
0.00/0.08	  map = \G0y1.G0(0) + y1G0(y1) 
0.00/0.08	  nil = 3 
0.00/0.08	  plus = \y0y1.2y0 
0.00/0.08	  s = \y0.1 + y0 
0.00/0.08	
0.00/0.08	Using this interpretation, the requirements translate to:
0.00/0.08	
0.00/0.08	  [[plus(0) _x0]] = x0 >= x0 = [[_x0]] 
0.00/0.08	  [[plus(s(_x0)) _x1]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[s(plus(_x0) _x1)]] 
0.00/0.08	  [[map(_F0) nil]] = 3 + F0(0) + 3F0(3) >= 3 = [[nil]] 
0.00/0.08	  [[map(_F0) cons(_x1, _x2)]] = 3 + x1 + x2 + F0(0) + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 3 + x1 + x2 + F0(0) + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0) _x2)]] 
0.00/0.09	  [[inc]] = \y0.3 + 3y0 > \y0.2 + 2y0 = [[map(plus(s(0)))]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  plus(s(X)) Y => s(plus(X) Y) 
0.00/0.09	  inc => map(plus(s(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	  plus(0, X) >? X 
0.00/0.09	  map(F, nil) >? nil 
0.00/0.09	  map(F, cons(X, Y)) >? cons(F X, map(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	  0 = 3 
0.00/0.09	  cons = \y0y1.3 + y0 + y1 
0.00/0.09	  map = \G0y1.3 + 3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) 
0.00/0.09	  nil = 0 
0.00/0.09	  plus = \y0y1.3 + y0 + y1 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[plus(0, _x0)]] = 6 + x0 > x0 = [[_x0]] 
0.00/0.09	  [[map(_F0, nil)]] = 3 + 4F0(0) > 0 = [[nil]] 
0.00/0.09	  [[map(_F0, cons(_x1, _x2))]] = 12 + 3x1 + 3x2 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) > 6 + x1 + 3x2 + F0(x1) + 2F0(0) + 2F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_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	  plus(0, X) => X 
0.00/0.09	  map(F, nil) => nil 
0.00/0.09	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
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