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	    0 : [] --> d 
0.00/0.09	    cons : [d * c] --> c 
0.00/0.09	    false : [] --> a 
0.00/0.09	    height : [] --> d -> d 
0.00/0.09	    if : [a * d] --> d 
0.00/0.09	    le : [d * d] --> a 
0.00/0.09	    map : [d -> d * c] --> c 
0.00/0.09	    maxlist : [d * c] --> d 
0.00/0.09	    nil : [] --> c 
0.00/0.09	    node : [b * c] --> d 
0.00/0.09	    s : [d] --> d 
0.00/0.09	    true : [] --> a 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
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	    le(0, x) => true 
0.00/0.09	    le(s(x), 0) => false 
0.00/0.09	    le(s(x), s(y)) => le(x, y) 
0.00/0.09	    maxlist(x, cons(y, z)) => if(le(x, y), maxlist(y, z)) 
0.00/0.09	    maxlist(x, nil) => x 
0.00/0.09	    height node(x, y) => s(maxlist(0, map(height, 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	  map(F, nil) >? nil 
0.00/0.09	  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
0.00/0.09	  le(0, X) >? true 
0.00/0.09	  le(s(X), 0) >? false 
0.00/0.09	  le(s(X), s(Y)) >? le(X, Y) 
0.00/0.09	  maxlist(X, cons(Y, Z)) >? if(le(X, Y), maxlist(Y, Z)) 
0.00/0.09	  maxlist(X, nil) >? X 
0.00/0.09	  height node(X, Y) >? s(maxlist(0, map(height, 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 = 0 
0.00/0.09	  cons = \y0y1.2 + 2y0 + 2y1 
0.00/0.09	  false = 0 
0.00/0.09	  height = \y0.0 
0.00/0.09	  if = \y0y1.y0 + y1 
0.00/0.09	  le = \y0y1.1 + y0 + y1 
0.00/0.09	  map = \G0y1.2y1 + G0(0) + 2y1G0(y1) 
0.00/0.09	  maxlist = \y0y1.1 + y0 + y1 
0.00/0.09	  nil = 0 
0.00/0.09	  node = \y0y1.3 + y0 + 3y1 
0.00/0.09	  s = \y0.y0 
0.00/0.09	  true = 0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[map(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
0.00/0.09	  [[map(_F0, cons(_x1, _x2))]] = 4 + 4x1 + 4x2 + F0(0) + 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 4F0(2 + 2x1 + 2x2) > 2 + 2x1 + 4x2 + 2F0(0) + 2F0(x1) + 4x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.09	  [[le(0, _x0)]] = 1 + x0 > 0 = [[true]] 
0.00/0.09	  [[le(s(_x0), 0)]] = 1 + x0 > 0 = [[false]] 
0.00/0.09	  [[le(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[le(_x0, _x1)]] 
0.00/0.09	  [[maxlist(_x0, cons(_x1, _x2))]] = 3 + x0 + 2x1 + 2x2 > 2 + x0 + x2 + 2x1 = [[if(le(_x0, _x1), maxlist(_x1, _x2))]] 
0.00/0.09	  [[maxlist(_x0, nil)]] = 1 + x0 > x0 = [[_x0]] 
0.00/0.09	  [[height node(_x0, _x1)]] = 3 + x0 + 3x1 > 1 + 2x1 = [[s(maxlist(0, map(height, _x1)))]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.09	  le(0, X) => true 
0.00/0.09	  le(s(X), 0) => false 
0.00/0.09	  maxlist(X, cons(Y, Z)) => if(le(X, Y), maxlist(Y, Z)) 
0.00/0.09	  maxlist(X, nil) => X 
0.00/0.09	  height node(X, Y) => s(maxlist(0, map(height, 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	  map(F, nil) >? nil 
0.00/0.09	  le(s(X), s(Y)) >? le(X, 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	  le = \y0y1.y0 + y1 
0.00/0.09	  map = \G0y1.3 + 3y1 + G0(0) 
0.00/0.09	  nil = 0 
0.00/0.09	  s = \y0.3 + 3y0 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.09	  [[le(s(_x0), s(_x1))]] = 6 + 3x0 + 3x1 > x0 + x1 = [[le(_x0, _x1)]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  map(F, nil) => nil 
0.00/0.09	  le(s(X), s(Y)) => le(X, 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