5.72/5.78	YES
5.72/5.79	We consider the system theBenchmark.
5.72/5.79	
5.72/5.79	  Alphabet:
5.72/5.79	
5.72/5.79	    0 : [] --> a 
5.72/5.79	    rec : [a -> b -> b * b * a] --> b 
5.72/5.79	    s : [a] --> a 
5.72/5.79	    xap : [a -> b -> b * a] --> b -> b 
5.72/5.79	    yap : [b -> b * b] --> b 
5.72/5.79	
5.72/5.79	  Rules:
5.72/5.79	
5.72/5.79	    rec(/\x./\y.yap(xap(f, x), y), z, 0) => z 
5.72/5.79	    rec(/\x./\y.yap(xap(f, x), y), z, s(u)) => yap(xap(f, s(u)), rec(/\v./\w.yap(xap(f, v), w), z, u)) 
5.72/5.79	    xap(f, x) => f x 
5.72/5.79	    yap(f, x) => f x 
5.72/5.79	
5.72/5.79	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
5.72/5.79	
5.72/5.79	Symbol xap 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:
5.72/5.79	
5.72/5.79	  Alphabet:
5.72/5.79	
5.72/5.79	    0 : [] --> a 
5.72/5.79	    rec : [a -> b -> b * b * a] --> b 
5.72/5.79	    s : [a] --> a 
5.72/5.79	    yap : [b -> b * b] --> b 
5.72/5.79	
5.72/5.79	  Rules:
5.72/5.79	
5.72/5.79	    rec(/\x./\y.yap(F(x), y), X, 0) => X 
5.72/5.79	    rec(/\x./\y.yap(F(x), y), X, s(Y)) => yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 
5.72/5.79	    yap(F, X) => F X 
5.72/5.79	
5.72/5.79	We use rule removal, following [Kop12, Theorem 2.23].
5.72/5.79	
5.72/5.79	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
5.72/5.79	
5.72/5.79	  rec(/\x./\y.yap(F(x), y), X, 0) >? X 
5.72/5.79	  rec(/\x./\y.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 
5.72/5.79	  yap(F, X) >? F X 
5.72/5.79	
5.72/5.79	We use a recursive path ordering as defined in [Kop12, Chapter 5].
5.72/5.79	
5.72/5.79	We choose Lex = {} and Mul = {0, @_{o -> o}, rec, s, yap}, and the following precedence: 0 > rec > s > yap > @_{o -> o}
5.72/5.79	
5.72/5.79	With these choices, we have:
5.72/5.79	
5.72/5.79	  1] rec(/\x./\y.yap(F(x), y), X, 0) > X  because [2], by definition 
5.72/5.79	  2] rec*(/\x./\y.yap(F(x), y), X, 0) >= X  because [3], by (Select) 
5.72/5.79	  3] X >= X  by (Meta) 
5.72/5.79	
5.72/5.79	  4] rec(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y))  because [5], by (Star) 
5.72/5.79	  5] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y))  because rec > yap, [6] and [14], by (Copy) 
5.72/5.79	  6] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= F(s(Y))  because [7], by (Select) 
5.72/5.79	  7] /\x.yap(F(rec*(/\y./\z.yap(F(y), z), X, s(Y))), x) >= F(s(Y))  because [8], by (Eta)[Kop13:2] 
5.72/5.79	  8] F(rec*(/\x./\y.yap(F(x), y), X, s(Y))) >= F(s(Y))  because [9], by (Meta) 
5.72/5.79	  9] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= s(Y)  because rec > s and [10], by (Copy) 
5.72/5.79	  10] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= Y  because [11], by (Select) 
5.72/5.79	  11] s(Y) >= Y  because [12], by (Star) 
5.72/5.79	  12] s*(Y) >= Y  because [13], by (Select) 
5.72/5.79	  13] Y >= Y  by (Meta) 
5.72/5.79	  14] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= rec(/\x./\y.yap(F(x), y), X, Y)  because rec in Mul, [15], [21] and [22], by (Stat) 
5.72/5.79	  15] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z)  because [16], by (Abs) 
5.72/5.79	  16] /\z.yap(F(y), z) >= /\z.yap(F(y), z)  because [17], by (Abs) 
5.72/5.79	  17] yap(F(y), x) >= yap(F(y), x)  because yap in Mul, [18] and [20], by (Fun) 
5.72/5.79	  18] F(y) >= F(y)  because [19], by (Meta) 
5.72/5.79	  19] y >= y  by (Var) 
5.72/5.79	  20] x >= x  by (Var) 
5.72/5.79	  21] X >= X  by (Meta) 
5.72/5.79	  22] s(Y) > Y  because [23], by definition 
5.72/5.79	  23] s*(Y) >= Y  because [13], by (Select) 
5.72/5.79	
5.72/5.79	  24] yap(F, X) > @_{o -> o}(F, X)  because [25], by definition 
5.72/5.79	  25] yap*(F, X) >= @_{o -> o}(F, X)  because yap > @_{o -> o}, [26] and [28], by (Copy) 
5.72/5.79	  26] yap*(F, X) >= F  because [27], by (Select) 
5.72/5.79	  27] F >= F  by (Meta) 
5.72/5.79	  28] yap*(F, X) >= X  because [29], by (Select) 
5.72/5.79	  29] X >= X  by (Meta) 
5.72/5.79	
5.72/5.79	We can thus remove the following rules:
5.72/5.79	
5.72/5.79	  rec(/\x./\y.yap(F(x), y), X, 0) => X 
5.72/5.79	  yap(F, X) => F X 
5.72/5.79	
5.72/5.79	We use rule removal, following [Kop12, Theorem 2.23].
5.72/5.79	
5.72/5.79	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
5.72/5.79	
5.72/5.79	  rec(/\x./\y.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 
5.72/5.79	
5.72/5.79	We orient these requirements with a polynomial interpretation in the natural numbers.
5.72/5.79	
5.72/5.79	The following interpretation satisfies the requirements:
5.72/5.79	
5.72/5.79	  rec = \G0y1y2.y1 + y2 + G0(0,0) + 2y2y2G0(y2,y2) + 2G0(y2,y1) 
5.72/5.79	  s = \y0.3 + 3y0 
5.72/5.79	  yap = \G0y1.y1 + 2G0(0) 
5.72/5.79	
5.72/5.79	Using this interpretation, the requirements translate to:
5.72/5.79	
5.72/5.79	  [[rec(/\x./\y.yap(_F0(x), y), _x1, s(_x2))]] = 57 + 3x1 + 54x2x2x2 + 162x2x2 + 165x2 + 2F0(0,0) + 36x2x2F0(3 + 3x2,0) + 40F0(3 + 3x2,0) + 72x2F0(3 + 3x2,0) > x2 + 2x2x2x2 + 3x1 + 2F0(0,0) + 2F0(3 + 3x2,0) + 4x2x2F0(x2,0) + 4F0(x2,0) = [[yap(_F0(s(_x2)), rec(/\x./\y.yap(_F0(x), y), _x1, _x2))]] 
5.72/5.79	
5.72/5.79	We can thus remove the following rules:
5.72/5.79	
5.72/5.79	  rec(/\x./\y.yap(F(x), y), X, s(Y)) => yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 
5.72/5.79	
5.72/5.79	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.
5.72/5.79	
5.72/5.79	
5.72/5.79	+++ Citations +++
5.72/5.79	
5.72/5.79	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
5.72/5.79	[Kop13:2]  C. Kop.  StarHorpo with an Eta-Rule.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/etahorpo.pdf, 2013.
5.72/5.79	EOF