0.05/0.60	YES
0.05/0.61	We consider the system theBenchmark.
0.05/0.61	
0.05/0.61	  Alphabet:
0.05/0.61	
0.05/0.61	    0 : [] --> a 
0.05/0.61	    rec : [] --> (a -> b -> b) -> b -> a -> b 
0.05/0.61	    s : [] --> a -> a 
0.05/0.61	
0.05/0.61	  Rules:
0.05/0.61	
0.05/0.61	    rec (/\x./\y.f x y) z 0 => z 
0.05/0.61	    rec (/\x./\y.f x y) z (s u) => f (s u) (rec (/\v./\w.f v w) z u) 
0.05/0.61	
0.05/0.61	Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term.
0.05/0.61	
0.05/0.61	We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0).  This leads to the following AFSM:
0.05/0.61	
0.05/0.61	  Alphabet:
0.05/0.61	
0.05/0.61	    0 : [] --> a 
0.05/0.61	    rec : [a -> b -> b * b * a] --> b 
0.05/0.61	    s : [a] --> a 
0.05/0.61	    ~AP1 : [a -> b -> b * a] --> b -> b 
0.05/0.61	
0.05/0.61	  Rules:
0.05/0.61	
0.05/0.61	    rec(/\x./\y.~AP1(F, x) y, X, 0) => X 
0.05/0.61	    rec(/\x./\y.~AP1(F, x) y, X, s(Y)) => ~AP1(F, s(Y)) rec(/\z./\u.~AP1(F, z) u, X, Y) 
0.05/0.61	    ~AP1(F, X) => F X 
0.05/0.61	
0.05/0.61	Symbol ~AP1 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:
0.05/0.61	
0.05/0.61	  Alphabet:
0.05/0.61	
0.05/0.61	    0 : [] --> a 
0.05/0.61	    rec : [a -> b -> b * b * a] --> b 
0.05/0.61	    s : [a] --> a 
0.05/0.61	
0.05/0.61	  Rules:
0.05/0.61	
0.05/0.61	    rec(/\x./\y.X(x, y), Y, 0) => Y 
0.05/0.61	    rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 
0.05/0.61	
0.05/0.61	We use rule removal, following [Kop12, Theorem 2.23].
0.05/0.61	
0.05/0.61	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.05/0.61	
0.05/0.61	  rec(/\x./\y.X(x, y), Y, 0) >? Y 
0.05/0.61	  rec(/\x./\y.X(x, y), Y, s(Z)) >? X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 
0.05/0.61	
0.05/0.61	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.05/0.61	
0.05/0.61	We choose Lex = {} and Mul = {0, rec, s}, and the following precedence: 0 > s > rec
0.05/0.61	
0.05/0.61	With these choices, we have:
0.05/0.61	
0.05/0.61	  1] rec(/\x./\y.X(x, y), Y, 0) >= Y  because [2], by (Star) 
0.05/0.61	  2] rec*(/\x./\y.X(x, y), Y, 0) >= Y  because [3], by (Select) 
0.05/0.61	  3] Y >= Y  by (Meta) 
0.05/0.61	
0.05/0.61	  4] rec(/\x./\y.X(x, y), Y, s(Z)) > X(s(Z), rec(/\x./\y.X(x, y), Y, Z))  because [5], by definition 
0.05/0.61	  5] rec*(/\x./\y.X(x, y), Y, s(Z)) >= X(s(Z), rec(/\x./\y.X(x, y), Y, Z))  because [6], by (Select) 
0.05/0.61	  6] X(rec*(/\x./\y.X(x, y), Y, s(Z)), rec*(/\z./\u.X(z, u), Y, s(Z))) >= X(s(Z), rec(/\x./\y.X(x, y), Y, Z))  because [7] and [10], by (Meta) 
0.05/0.61	  7] rec*(/\x./\y.X(x, y), Y, s(Z)) >= s(Z)  because [8], by (Select) 
0.05/0.61	  8] s(Z) >= s(Z)  because s in Mul and [9], by (Fun) 
0.05/0.61	  9] Z >= Z  by (Meta) 
0.05/0.61	  10] rec*(/\x./\y.X(x, y), Y, s(Z)) >= rec(/\x./\y.X(x, y), Y, Z)  because rec in Mul, [11], [16] and [17], by (Stat) 
0.05/0.61	  11] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [12], by (Abs) 
0.05/0.61	  12] /\z.X(y, z) >= /\z.X(y, z)  because [13], by (Abs) 
0.05/0.61	  13] X(y, x) >= X(y, x)  because [14] and [15], by (Meta) 
0.05/0.61	  14] y >= y  by (Var) 
0.05/0.61	  15] x >= x  by (Var) 
0.05/0.61	  16] Y >= Y  by (Meta) 
0.05/0.61	  17] s(Z) > Z  because [18], by definition 
0.05/0.61	  18] s*(Z) >= Z  because [9], by (Select) 
0.05/0.61	
0.05/0.61	We can thus remove the following rules:
0.05/0.61	
0.05/0.61	  rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 
0.05/0.61	
0.05/0.61	We use rule removal, following [Kop12, Theorem 2.23].
0.05/0.61	
0.05/0.61	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.05/0.61	
0.05/0.61	  rec(/\x./\y.X(x, y), Y, 0) >? Y 
0.05/0.61	
0.05/0.61	We orient these requirements with a polynomial interpretation in the natural numbers.
0.05/0.61	
0.05/0.61	The following interpretation satisfies the requirements:
0.05/0.61	
0.05/0.61	  0 = 3 
0.05/0.61	  rec = \G0y1y2.3 + y1 + y2 + G0(0,0) 
0.05/0.61	
0.05/0.61	Using this interpretation, the requirements translate to:
0.05/0.61	
0.05/0.61	  [[rec(/\x./\y._x0(x, y), _x1, 0)]] = 6 + x1 + F0(0,0) > x1 = [[_x1]] 
0.05/0.61	
0.05/0.61	We can thus remove the following rules:
0.05/0.61	
0.05/0.61	  rec(/\x./\y.X(x, y), Y, 0) => Y 
0.05/0.61	
0.05/0.61	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.
0.05/0.61	
0.05/0.61	
0.05/0.61	+++ Citations +++
0.05/0.61	
0.05/0.61	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.05/0.61	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.05/0.61	EOF