4.53/4.58	YES
4.53/4.60	We consider the system theBenchmark.
4.53/4.60	
4.53/4.60	  Alphabet:
4.53/4.60	
4.53/4.60	    0 : [] --> c 
4.53/4.60	    1 : [] --> c 
4.53/4.60	    add : [] --> c -> a -> c 
4.53/4.60	    cons : [a * b] --> b 
4.53/4.60	    fold : [c -> a -> c * b * c] --> c 
4.53/4.60	    mul : [] --> c -> a -> c 
4.53/4.60	    nil : [] --> b 
4.53/4.60	    prod : [b] --> c 
4.53/4.60	    sum : [b] --> c 
4.53/4.60	    xap : [c -> a -> c * c] --> a -> c 
4.53/4.60	    yap : [a -> c * a] --> c 
4.53/4.60	
4.53/4.60	  Rules:
4.53/4.60	
4.53/4.60	    fold(/\x./\y.yap(xap(f, x), y), nil, z) => z 
4.53/4.60	    fold(/\x./\y.yap(xap(f, x), y), cons(z, u), v) => fold(/\w./\x'.yap(xap(f, w), x'), u, yap(xap(f, v), z)) 
4.53/4.60	    sum(x) => fold(/\y./\z.yap(xap(add, y), z), x, 0) 
4.53/4.60	    fold(mul, x, 1) => prod(x) 
4.53/4.60	    xap(f, x) => f x 
4.53/4.60	    yap(f, x) => f x 
4.53/4.60	
4.53/4.60	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
4.53/4.60	
4.53/4.60	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:
4.53/4.60	
4.53/4.60	  Alphabet:
4.53/4.60	
4.53/4.60	    0 : [] --> c 
4.53/4.60	    1 : [] --> c 
4.53/4.60	    add : [c] --> a -> c 
4.53/4.60	    cons : [a * b] --> b 
4.53/4.60	    fold : [c -> a -> c * b * c] --> c 
4.53/4.60	    mul : [] --> c -> a -> c 
4.53/4.60	    nil : [] --> b 
4.53/4.60	    prod : [b] --> c 
4.53/4.60	    sum : [b] --> c 
4.53/4.60	    yap : [a -> c * a] --> c 
4.53/4.60	
4.53/4.60	  Rules:
4.53/4.60	
4.53/4.60	    fold(/\x./\y.yap(F(x), y), nil, X) => X 
4.53/4.60	    fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) => fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
4.53/4.60	    sum(X) => fold(/\x./\y.yap(add(x), y), X, 0) 
4.53/4.60	    fold(mul, X, 1) => prod(X) 
4.53/4.60	    yap(F, X) => F X 
4.53/4.60	
4.53/4.60	We use rule removal, following [Kop12, Theorem 2.23].
4.53/4.60	
4.53/4.60	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
4.53/4.60	
4.53/4.60	  fold(/\x./\y.yap(F(x), y), nil, X) >? X 
4.53/4.60	  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) >? fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
4.53/4.60	  sum(X) >? fold(/\x./\y.yap(add(x), y), X, 0) 
4.53/4.60	  fold(mul, X, 1) >? prod(X) 
4.53/4.60	  yap(F, X) >? F X 
4.53/4.60	
4.53/4.60	We use a recursive path ordering as defined in [Kop12, Chapter 5].
4.53/4.60	
4.53/4.60	Argument functions:
4.53/4.60	
4.53/4.60	  [[0]] = _|_ 
4.53/4.60	  [[fold(x_1, x_2, x_3)]] = fold(x_2, x_3, x_1) 
4.53/4.60	  [[prod(x_1)]] = x_1 
4.53/4.60	
4.53/4.60	We choose Lex = {fold} and Mul = {1, @_{o -> o}, add, cons, mul, nil, sum, yap}, and the following precedence: 1 > mul > nil > cons > sum > add > fold > yap > @_{o -> o}
4.53/4.60	
4.53/4.60	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
4.53/4.60	
4.53/4.60	  fold(/\x./\y.yap(F(x), y), nil, X) >= X 
4.53/4.60	  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) > fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) 
4.53/4.60	  sum(X) >= fold(/\x./\y.yap(add(x), y), X, _|_) 
4.53/4.60	  fold(mul, X, 1) > X 
4.53/4.60	  yap(F, X) >= @_{o -> o}(F, X) 
4.53/4.60	
4.53/4.60	With these choices, we have:
4.53/4.60	
4.53/4.60	  1] fold(/\x./\y.yap(F(x), y), nil, X) >= X  because [2], by (Star) 
4.53/4.60	  2] fold*(/\x./\y.yap(F(x), y), nil, X) >= X  because [3], by (Select) 
4.53/4.60	  3] X >= X  by (Meta) 
4.53/4.60	
4.53/4.60	  4] fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) > fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X))  because [5], by definition 
4.53/4.60	  5] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X))  because [6], by (Select) 
4.53/4.60	  6] yap(F(fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z)), fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X))  because [7], by (Star) 
4.53/4.60	  7] yap*(F(fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z)), fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X))  because [8], by (Select) 
4.53/4.60	  8] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X))  because [9], [12], [19] and [21], by (Stat) 
4.53/4.60	  9] cons(X, Y) > Y  because [10], by definition 
4.53/4.60	  10] cons*(X, Y) >= Y  because [11], by (Select) 
4.53/4.60	  11] Y >= Y  by (Meta) 
4.53/4.60	  12] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= /\x./\y.yap(F(x), y)  because [13], by (Select) 
4.53/4.60	  13] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z)  because [14], by (Abs) 
4.53/4.60	  14] /\z.yap(F(y), z) >= /\z.yap(F(y), z)  because [15], by (Abs) 
4.53/4.60	  15] yap(F(y), x) >= yap(F(y), x)  because yap in Mul, [16] and [18], by (Fun) 
4.53/4.60	  16] F(y) >= F(y)  because [17], by (Meta) 
4.53/4.60	  17] y >= y  by (Var) 
4.53/4.60	  18] x >= x  by (Var) 
4.53/4.60	  19] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= Y  because [20], by (Select) 
4.53/4.60	  20] cons(X, Y) >= Y  because [10], by (Star) 
4.53/4.60	  21] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= yap(F(Z), X)  because fold > yap, [22] and [27], by (Copy) 
4.53/4.60	  22] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= F(Z)  because [23], by (Select) 
4.53/4.60	  23] /\z.yap(F(fold*(/\u./\v.yap(F(u), v), cons(X, Y), Z)), z) >= F(Z)  because [24], by (Eta)[Kop13:2] 
4.53/4.60	  24] F(fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= F(Z)  because [25], by (Meta) 
4.53/4.60	  25] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= Z  because [26], by (Select) 
4.53/4.60	  26] Z >= Z  by (Meta) 
4.53/4.60	  27] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= X  because [28], by (Select) 
4.53/4.60	  28] cons(X, Y) >= X  because [29], by (Star) 
4.53/4.60	  29] cons*(X, Y) >= X  because [30], by (Select) 
4.53/4.60	  30] X >= X  by (Meta) 
4.53/4.60	
4.53/4.60	  31] sum(X) >= fold(/\x./\y.yap(add(x), y), X, _|_)  because [32], by (Star) 
4.53/4.60	  32] sum*(X) >= fold(/\x./\y.yap(add(x), y), X, _|_)  because sum > fold, [33], [41] and [43], by (Copy) 
4.53/4.60	  33] sum*(X) >= /\y./\z.yap(add(y), z)  because [34], by (F-Abs) 
4.53/4.60	  34] sum*(X, x) >= /\z.yap(add(x), z)  because [35], by (F-Abs) 
4.53/4.60	  35] sum*(X, x, y) >= yap(add(x), y)  because sum > yap, [36] and [39], by (Copy) 
4.53/4.60	  36] sum*(X, x, y) >= add(x)  because sum > add and [37], by (Copy) 
4.53/4.60	  37] sum*(X, x, y) >= x  because [38], by (Select) 
4.53/4.60	  38] x >= x  by (Var) 
4.53/4.60	  39] sum*(X, x, y) >= y  because [40], by (Select) 
4.53/4.60	  40] y >= y  by (Var) 
4.53/4.60	  41] sum*(X) >= X  because [42], by (Select) 
4.53/4.60	  42] X >= X  by (Meta) 
4.53/4.60	  43] sum*(X) >= _|_  by (Bot) 
4.53/4.60	
4.53/4.60	  44] fold(mul, X, 1) > X  because [45], by definition 
4.53/4.60	  45] fold*(mul, X, 1) >= X  because [46], by (Select) 
4.53/4.60	  46] X >= X  by (Meta) 
4.53/4.60	
4.53/4.60	  47] yap(F, X) >= @_{o -> o}(F, X)  because [48], by (Star) 
4.53/4.60	  48] yap*(F, X) >= @_{o -> o}(F, X)  because yap > @_{o -> o}, [49] and [51], by (Copy) 
4.53/4.60	  49] yap*(F, X) >= F  because [50], by (Select) 
4.53/4.60	  50] F >= F  by (Meta) 
4.53/4.60	  51] yap*(F, X) >= X  because [52], by (Select) 
4.53/4.60	  52] X >= X  by (Meta) 
4.53/4.60	
4.53/4.60	We can thus remove the following rules:
4.53/4.60	
4.53/4.60	  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) => fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
4.53/4.60	  fold(mul, X, 1) => prod(X) 
4.53/4.60	
4.53/4.60	We use rule removal, following [Kop12, Theorem 2.23].
4.53/4.60	
4.53/4.60	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
4.53/4.60	
4.53/4.60	  fold(/\x./\y.yap(F(x), y), nil, X) >? X 
4.53/4.60	  sum(X) >? fold(/\x./\y.yap(add(x), y), X, 0) 
4.53/4.60	  yap(F, X) >? F X 
4.53/4.60	
4.53/4.60	We orient these requirements with a polynomial interpretation in the natural numbers.
4.53/4.60	
4.53/4.60	The following interpretation satisfies the requirements:
4.53/4.60	
4.53/4.60	  0 = 0 
4.53/4.60	  add = \y0y1.y0 
4.53/4.60	  fold = \G0y1y2.y1 + y2 + G0(0,0) 
4.53/4.60	  nil = 3 
4.53/4.60	  sum = \y0.3 + 3y0 
4.53/4.60	  yap = \G0y1.y1 + G0(y1) 
4.53/4.60	
4.53/4.60	Using this interpretation, the requirements translate to:
4.53/4.60	
4.53/4.60	  [[fold(/\x./\y.yap(_F0(x), y), nil, _x1)]] = 3 + x1 + F0(0,0) > x1 = [[_x1]] 
4.53/4.60	  [[sum(_x0)]] = 3 + 3x0 > x0 = [[fold(/\x./\y.yap(add(x), y), _x0, 0)]] 
4.53/4.60	  [[yap(_F0, _x1)]] = x1 + F0(x1) >= x1 + F0(x1) = [[_F0 _x1]] 
4.53/4.60	
4.53/4.60	We can thus remove the following rules:
4.53/4.60	
4.53/4.60	  fold(/\x./\y.yap(F(x), y), nil, X) => X 
4.53/4.60	  sum(X) => fold(/\x./\y.yap(add(x), y), X, 0) 
4.53/4.60	
4.53/4.60	We use rule removal, following [Kop12, Theorem 2.23].
4.53/4.60	
4.53/4.60	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
4.53/4.60	
4.53/4.60	  yap(F, X) >? F X 
4.53/4.60	
4.53/4.60	We orient these requirements with a polynomial interpretation in the natural numbers.
4.53/4.60	
4.53/4.60	The following interpretation satisfies the requirements:
4.53/4.60	
4.53/4.60	  yap = \G0y1.1 + y1 + G0(y1) 
4.53/4.60	
4.53/4.60	Using this interpretation, the requirements translate to:
4.53/4.60	
4.53/4.60	  [[yap(_F0, _x1)]] = 1 + x1 + F0(x1) > x1 + F0(x1) = [[_F0 _x1]] 
4.53/4.60	
4.53/4.60	We can thus remove the following rules:
4.53/4.60	
4.53/4.60	  yap(F, X) => F X 
4.53/4.60	
4.53/4.60	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.
4.53/4.60	
4.53/4.60	
4.53/4.60	+++ Citations +++
4.53/4.60	
4.53/4.60	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
4.53/4.60	[Kop13:2]  C. Kop.  StarHorpo with an Eta-Rule.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/etahorpo.pdf, 2013.
4.53/4.60	EOF