0.00/0.09	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 : [] --> nat 
0.00/0.09	    rec : [] --> nat -> a -> (nat -> a -> a) -> a 
0.00/0.09	    s : [] --> nat -> nat 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    rec 0 x (/\y./\z.f y z) => x 
0.00/0.09	    rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) 
0.00/0.09	
0.00/0.09	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.
0.00/0.09	
0.00/0.09	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:
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    0 : [] --> nat 
0.00/0.09	    rec : [nat * a * nat -> a -> a] --> a 
0.00/0.09	    s : [nat] --> nat 
0.00/0.09	    ~AP1 : [nat -> a -> a * nat] --> a -> a 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    rec(0, X, /\x./\y.~AP1(F, x) y) => X 
0.00/0.09	    rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) 
0.00/0.09	    ~AP1(F, X) => F X 
0.00/0.09	
0.00/0.09	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:
0.00/0.09	
0.00/0.09	  Alphabet:
0.00/0.09	
0.00/0.09	    0 : [] --> nat 
0.00/0.09	    rec : [nat * a * nat -> a -> a] --> a 
0.00/0.09	    s : [nat] --> nat 
0.00/0.09	
0.00/0.09	  Rules:
0.00/0.09	
0.00/0.09	    rec(0, X, /\x./\y.Y(x, y)) => X 
0.00/0.09	    rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 
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	  rec(0, X, /\x./\y.Y(x, y)) >? X 
0.00/0.09	  rec(s(X), Y, /\x./\y.Z(x, y)) >? Z(X, rec(X, Y, /\z./\u.Z(z, u))) 
0.00/0.09	
0.00/0.09	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.09	
0.00/0.09	We choose Lex = {} and Mul = {0, rec, s}, and the following precedence: 0 > rec > s
0.00/0.09	
0.00/0.09	With these choices, we have:
0.00/0.09	
0.00/0.09	  1] rec(0, X, /\x./\y.Y(x, y)) >= X  because [2], by (Star) 
0.00/0.09	  2] rec*(0, X, /\x./\y.Y(x, y)) >= X  because [3], by (Select) 
0.00/0.09	  3] X >= X  by (Meta) 
0.00/0.09	
0.00/0.09	  4] rec(s(X), Y, /\x./\y.Z(x, y)) > Z(X, rec(X, Y, /\x./\y.Z(x, y)))  because [5], by definition 
0.00/0.09	  5] rec*(s(X), Y, /\x./\y.Z(x, y)) >= Z(X, rec(X, Y, /\x./\y.Z(x, y)))  because [6], by (Select) 
0.00/0.09	  6] Z(rec*(s(X), Y, /\x./\y.Z(x, y)), rec*(s(X), Y, /\z./\u.Z(z, u))) >= Z(X, rec(X, Y, /\x./\y.Z(x, y)))  because [7] and [11], by (Meta) 
0.00/0.09	  7] rec*(s(X), Y, /\x./\y.Z(x, y)) >= X  because [8], by (Select) 
0.00/0.09	  8] s(X) >= X  because [9], by (Star) 
0.00/0.09	  9] s*(X) >= X  because [10], by (Select) 
0.00/0.09	  10] X >= X  by (Meta) 
0.00/0.09	  11] rec*(s(X), Y, /\x./\y.Z(x, y)) >= rec(X, Y, /\x./\y.Z(x, y))  because rec in Mul, [12], [14] and [15], by (Stat) 
0.00/0.09	  12] s(X) > X  because [13], by definition 
0.00/0.09	  13] s*(X) >= X  because [10], by (Select) 
0.00/0.09	  14] Y >= Y  by (Meta) 
0.00/0.09	  15] /\x./\z.Z(x, z) >= /\x./\z.Z(x, z)  because [16], by (Abs) 
0.00/0.09	  16] /\z.Z(y, z) >= /\z.Z(y, z)  because [17], by (Abs) 
0.00/0.09	  17] Z(y, x) >= Z(y, x)  because [18] and [19], by (Meta) 
0.00/0.09	  18] y >= y  by (Var) 
0.00/0.09	  19] x >= x  by (Var) 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 
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	  rec(0, X, /\x./\y.Y(x, y)) >? X 
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 = 3 
0.00/0.09	  rec = \y0y1G2.3 + y0 + y1 + G2(0,0) 
0.00/0.09	
0.00/0.09	Using this interpretation, the requirements translate to:
0.00/0.09	
0.00/0.09	  [[rec(0, _x0, /\x./\y._x1(x, y))]] = 6 + x0 + F1(0,0) > x0 = [[_x0]] 
0.00/0.09	
0.00/0.09	We can thus remove the following rules:
0.00/0.09	
0.00/0.09	  rec(0, X, /\x./\y.Y(x, y)) => X 
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	[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
0.00/0.09	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.09	EOF