0.00/0.10	YES
0.00/0.11	We consider the system theBenchmark.
0.00/0.11	
0.00/0.11	  Alphabet:
0.00/0.11	
0.00/0.11	    append : [a * a] --> a 
0.00/0.11	    combine : [a * a] --> a 
0.00/0.11	    cons : [a * a] --> a 
0.00/0.11	    levels : [] --> a -> a 
0.00/0.11	    map : [a -> a * a] --> a 
0.00/0.11	    nil : [] --> a 
0.00/0.11	    node : [a * a] --> a 
0.00/0.11	    zip : [a * a] --> a 
0.00/0.11	
0.00/0.11	  Rules:
0.00/0.11	
0.00/0.11	    map(f, nil) => nil 
0.00/0.11	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
0.00/0.11	    append(x, nil) => x 
0.00/0.11	    append(nil, x) => x 
0.00/0.11	    append(cons(x, y), z) => cons(x, append(y, z)) 
0.00/0.11	    zip(nil, x) => x 
0.00/0.11	    zip(x, nil) => x 
0.00/0.11	    zip(cons(x, y), cons(z, u)) => cons(append(x, z), zip(y, u)) 
0.00/0.11	    combine(x, nil) => x 
0.00/0.11	    combine(x, cons(y, z)) => combine(zip(x, y), z) 
0.00/0.11	    levels node(x, y) => cons(cons(x, nil), combine(nil, map(levels, y))) 
0.00/0.11	
0.00/0.11	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.11	
0.00/0.11	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.11	
0.00/0.11	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.11	
0.00/0.11	  map(F, nil) >? nil 
0.00/0.11	  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
0.00/0.11	  append(X, nil) >? X 
0.00/0.11	  append(nil, X) >? X 
0.00/0.11	  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
0.00/0.11	  zip(nil, X) >? X 
0.00/0.11	  zip(X, nil) >? X 
0.00/0.11	  zip(cons(X, Y), cons(Z, U)) >? cons(append(X, Z), zip(Y, U)) 
0.00/0.11	  combine(X, nil) >? X 
0.00/0.11	  combine(X, cons(Y, Z)) >? combine(zip(X, Y), Z) 
0.00/0.11	  levels node(X, Y) >? cons(cons(X, nil), combine(nil, map(levels, Y))) 
0.00/0.11	
0.00/0.11	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.11	
0.00/0.11	The following interpretation satisfies the requirements:
0.00/0.11	
0.00/0.11	  append = \y0y1.y0 + y1 
0.00/0.11	  combine = \y0y1.y0 + y1 
0.00/0.11	  cons = \y0y1.1 + y0 + y1 
0.00/0.11	  levels = \y0.0 
0.00/0.11	  map = \G0y1.2y1 + 2y1G0(y1) + 2G0(0) 
0.00/0.11	  nil = 0 
0.00/0.11	  node = \y0y1.3 + 3y0 + 3y1 
0.00/0.11	  zip = \y0y1.y0 + y1 
0.00/0.11	
0.00/0.11	Using this interpretation, the requirements translate to:
0.00/0.11	
0.00/0.11	  [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
0.00/0.11	  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(0) + 2F0(1 + x1 + x2) > 1 + x1 + 2x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[cons(_F0 _x1, map(_F0, _x2))]] 
0.00/0.11	  [[append(_x0, nil)]] = x0 >= x0 = [[_x0]] 
0.00/0.11	  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
0.00/0.11	  [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
0.00/0.11	  [[zip(nil, _x0)]] = x0 >= x0 = [[_x0]] 
0.00/0.11	  [[zip(_x0, nil)]] = x0 >= x0 = [[_x0]] 
0.00/0.11	  [[zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2 + x0 + x1 + x2 + x3 > 1 + x0 + x1 + x2 + x3 = [[cons(append(_x0, _x2), zip(_x1, _x3))]] 
0.00/0.11	  [[combine(_x0, nil)]] = x0 >= x0 = [[_x0]] 
0.00/0.11	  [[combine(_x0, cons(_x1, _x2))]] = 1 + x0 + x1 + x2 > x0 + x1 + x2 = [[combine(zip(_x0, _x1), _x2)]] 
0.00/0.11	  [[levels node(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x0 + 2x1 = [[cons(cons(_x0, nil), combine(nil, map(levels, _x1)))]] 
0.00/0.11	
0.00/0.11	We can thus remove the following rules:
0.00/0.11	
0.00/0.11	  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
0.00/0.11	  zip(cons(X, Y), cons(Z, U)) => cons(append(X, Z), zip(Y, U)) 
0.00/0.11	  combine(X, cons(Y, Z)) => combine(zip(X, Y), Z) 
0.00/0.11	  levels node(X, Y) => cons(cons(X, nil), combine(nil, map(levels, Y))) 
0.00/0.11	
0.00/0.11	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.11	
0.00/0.11	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.11	
0.00/0.11	  map(F, nil) >? nil 
0.00/0.11	  append(X, nil) >? X 
0.00/0.11	  append(nil, X) >? X 
0.00/0.11	  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
0.00/0.11	  zip(nil, X) >? X 
0.00/0.11	  zip(X, nil) >? X 
0.00/0.11	  combine(X, nil) >? X 
0.00/0.11	
0.00/0.11	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.11	
0.00/0.11	The following interpretation satisfies the requirements:
0.00/0.11	
0.00/0.11	  append = \y0y1.3 + y1 + 3y0 
0.00/0.11	  combine = \y0y1.3 + y0 + y1 
0.00/0.11	  cons = \y0y1.3 + y0 + y1 
0.00/0.11	  map = \G0y1.3 + 3y1 + G0(0) 
0.00/0.11	  nil = 0 
0.00/0.11	  zip = \y0y1.3 + y0 + y1 
0.00/0.11	
0.00/0.11	Using this interpretation, the requirements translate to:
0.00/0.11	
0.00/0.11	  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.11	  [[append(_x0, nil)]] = 3 + 3x0 > x0 = [[_x0]] 
0.00/0.11	  [[append(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 
0.00/0.11	  [[append(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 
0.00/0.11	  [[zip(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 
0.00/0.11	  [[zip(_x0, nil)]] = 3 + x0 > x0 = [[_x0]] 
0.00/0.11	  [[combine(_x0, nil)]] = 3 + x0 > x0 = [[_x0]] 
0.00/0.11	
0.00/0.11	We can thus remove the following rules:
0.00/0.11	
0.00/0.11	  map(F, nil) => nil 
0.00/0.11	  append(X, nil) => X 
0.00/0.11	  append(nil, X) => X 
0.00/0.11	  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
0.00/0.11	  zip(nil, X) => X 
0.00/0.11	  zip(X, nil) => X 
0.00/0.11	  combine(X, nil) => X 
0.00/0.11	
0.00/0.11	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.
0.00/0.11	
0.00/0.11	
0.00/0.11	+++ Citations +++
0.00/0.11	
0.00/0.11	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.11	EOF