0.00/0.05	YES
0.00/0.05	We consider the system theBenchmark.
0.00/0.05	
0.00/0.05	  Alphabet:
0.00/0.05	
0.00/0.05	    0 : [] --> nat 
0.00/0.05	    bool : [nat] --> boolean 
0.00/0.05	    cons : [nat * list] --> list 
0.00/0.05	    consif : [boolean * nat * list] --> list 
0.00/0.05	    false : [] --> boolean 
0.00/0.05	    filter : [nat -> boolean * list] --> list 
0.00/0.05	    nil : [] --> list 
0.00/0.05	    rand : [nat] --> nat 
0.00/0.05	    s : [nat] --> nat 
0.00/0.05	    true : [] --> boolean 
0.00/0.05	
0.00/0.05	  Rules:
0.00/0.05	
0.00/0.05	    rand(x) => x 
0.00/0.05	    rand(s(x)) => rand(x) 
0.00/0.05	    bool(0) => false 
0.00/0.05	    bool(s(0)) => true 
0.00/0.05	    filter(f, nil) => nil 
0.00/0.05	    filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) 
0.00/0.05	    consif(true, x, y) => cons(x, y) 
0.00/0.05	    consif(false, x, y) => y 
0.00/0.05	
0.00/0.05	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.05	
0.00/0.05	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.05	
0.00/0.05	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.05	
0.00/0.05	  rand(X) >? X 
0.00/0.05	  rand(s(X)) >? rand(X) 
0.00/0.05	  bool(0) >? false 
0.00/0.05	  bool(s(0)) >? true 
0.00/0.05	  filter(F, nil) >? nil 
0.00/0.05	  filter(F, cons(X, Y)) >? consif(F X, X, filter(F, Y)) 
0.00/0.05	  consif(true, X, Y) >? cons(X, Y) 
0.00/0.05	  consif(false, X, Y) >? Y 
0.00/0.05	
0.00/0.05	We orient these requirements with a polynomial interpretation in the natural numbers.
0.00/0.05	
0.00/0.05	The following interpretation satisfies the requirements:
0.00/0.05	
0.00/0.05	  0 = 3 
0.00/0.05	  bool = \y0.3 + 3y0 
0.00/0.05	  cons = \y0y1.2 + 2y0 + 2y1 
0.00/0.05	  consif = \y0y1y2.1 + y0 + 2y1 + 2y2 
0.00/0.05	  false = 1 
0.00/0.05	  filter = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 
0.00/0.05	  nil = 2 
0.00/0.05	  rand = \y0.3 + 2y0 
0.00/0.05	  s = \y0.3 + 3y0 
0.00/0.05	  true = 3 
0.00/0.05	
0.00/0.05	Using this interpretation, the requirements translate to:
0.00/0.05	
0.00/0.05	  [[rand(_x0)]] = 3 + 2x0 > x0 = [[_x0]] 
0.00/0.05	  [[rand(s(_x0))]] = 9 + 6x0 > 3 + 2x0 = [[rand(_x0)]] 
0.00/0.05	  [[bool(0)]] = 12 > 1 = [[false]] 
0.00/0.05	  [[bool(s(0))]] = 39 > 3 = [[true]] 
0.00/0.05	  [[filter(_F0, nil)]] = 4 + 5F0(2) > 2 = [[nil]] 
0.00/0.05	  [[filter(_F0, cons(_x1, _x2))]] = 4 + 4x1 + 4x2 + 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 5F0(2 + 2x1 + 2x2) > 1 + 3x1 + 4x2 + F0(x1) + 2F0(x2) + 4x2F0(x2) = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] 
0.00/0.05	  [[consif(true, _x0, _x1)]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[cons(_x0, _x1)]] 
0.00/0.05	  [[consif(false, _x0, _x1)]] = 2 + 2x0 + 2x1 > x1 = [[_x1]] 
0.00/0.05	
0.00/0.05	We can thus remove the following rules:
0.00/0.05	
0.00/0.05	  rand(X) => X 
0.00/0.05	  rand(s(X)) => rand(X) 
0.00/0.05	  bool(0) => false 
0.00/0.05	  bool(s(0)) => true 
0.00/0.05	  filter(F, nil) => nil 
0.00/0.05	  filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 
0.00/0.05	  consif(true, X, Y) => cons(X, Y) 
0.00/0.05	  consif(false, X, Y) => Y 
0.00/0.05	
0.00/0.05	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.
0.00/0.05	
0.00/0.05	
0.00/0.05	+++ Citations +++
0.00/0.05	
0.00/0.05	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.05	EOF