0.00/0.11	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 : [c * c] --> c 
0.00/0.11	    cons : [b * c] --> c 
0.00/0.11	    flatwith : [a -> b * b] --> c 
0.00/0.11	    flatwithsub : [a -> b * c] --> c 
0.00/0.11	    leaf : [a] --> b 
0.00/0.11	    nil : [] --> c 
0.00/0.11	    node : [c] --> b 
0.00/0.11	
0.00/0.11	  Rules:
0.00/0.11	
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	    flatwith(f, leaf(x)) => cons(f x, nil) 
0.00/0.11	    flatwith(f, node(x)) => flatwithsub(f, x) 
0.00/0.11	    flatwithsub(f, nil) => nil 
0.00/0.11	    flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(f, 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	  append(nil, X) >? X 
0.00/0.11	  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
0.00/0.11	  flatwith(F, leaf(X)) >? cons(F X, nil) 
0.00/0.11	  flatwith(F, node(X)) >? flatwithsub(F, X) 
0.00/0.11	  flatwithsub(F, nil) >? nil 
0.00/0.11	  flatwithsub(F, cons(X, Y)) >? append(flatwith(F, X), flatwithsub(F, 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	  cons = \y0y1.1 + y0 + y1 
0.00/0.11	  flatwith = \G0y1.y1 + G0(y1) + 2y1G0(y1) 
0.00/0.11	  flatwithsub = \G0y1.y1 + G0(0) + G0(y1) + 2y1G0(y1) 
0.00/0.11	  leaf = \y0.3 + 3y0 
0.00/0.11	  nil = 2 
0.00/0.11	  node = \y0.3 + 3y0 
0.00/0.11	
0.00/0.11	Using this interpretation, the requirements translate to:
0.00/0.11	
0.00/0.11	  [[append(nil, _x0)]] = 2 + 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	  [[flatwith(_F0, leaf(_x1))]] = 3 + 3x1 + 6x1F0(3 + 3x1) + 7F0(3 + 3x1) >= 3 + x1 + F0(x1) = [[cons(_F0 _x1, nil)]] 
0.00/0.11	  [[flatwith(_F0, node(_x1))]] = 3 + 3x1 + 6x1F0(3 + 3x1) + 7F0(3 + 3x1) > x1 + F0(0) + F0(x1) + 2x1F0(x1) = [[flatwithsub(_F0, _x1)]] 
0.00/0.11	  [[flatwithsub(_F0, nil)]] = 2 + F0(0) + 5F0(2) >= 2 = [[nil]] 
0.00/0.11	  [[flatwithsub(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > x1 + x2 + F0(0) + F0(x1) + F0(x2) + 2x1F0(x1) + 2x2F0(x2) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 
0.00/0.11	
0.00/0.11	We can thus remove the following rules:
0.00/0.11	
0.00/0.11	  append(nil, X) => X 
0.00/0.11	  flatwith(F, node(X)) => flatwithsub(F, X) 
0.00/0.11	  flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(F, 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	  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
0.00/0.11	  flatwith(F, leaf(X)) >? cons(F X, nil) 
0.00/0.11	  flatwithsub(F, nil) >? nil 
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.y1 + 3y0 
0.00/0.11	  cons = \y0y1.y0 + y1 
0.00/0.11	  flatwith = \G0y1.3 + 3y1 + G0(0) + y1G0(y1) 
0.00/0.11	  flatwithsub = \G0y1.3 + 3y1 + G0(0) 
0.00/0.11	  leaf = \y0.3 + 3y0 
0.00/0.11	  nil = 0 
0.00/0.11	
0.00/0.11	Using this interpretation, the requirements translate to:
0.00/0.11	
0.00/0.11	  [[append(cons(_x0, _x1), _x2)]] = x2 + 3x0 + 3x1 >= x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 
0.00/0.11	  [[flatwith(_F0, leaf(_x1))]] = 12 + 9x1 + F0(0) + 3x1F0(3 + 3x1) + 3F0(3 + 3x1) > x1 + F0(x1) = [[cons(_F0 _x1, nil)]] 
0.00/0.11	  [[flatwithsub(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 
0.00/0.11	
0.00/0.11	We can thus remove the following rules:
0.00/0.11	
0.00/0.11	  flatwith(F, leaf(X)) => cons(F X, nil) 
0.00/0.11	  flatwithsub(F, nil) => nil 
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	  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
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.y1 + 3y0 
0.00/0.11	  cons = \y0y1.1 + 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	  [[append(cons(_x0, _x1), _x2)]] = 3 + x2 + 3x0 + 3x1 > 1 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 
0.00/0.11	
0.00/0.11	We can thus remove the following rules:
0.00/0.11	
0.00/0.11	  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
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