We consider the system h03.

  Alphabet:

    0 : [] --> c 
    1 : [] --> c 
    add : [] --> c -> a -> c 
    cons : [a * b] --> b 
    fold : [c -> a -> c * b * c] --> c 
    mul : [] --> c -> a -> c 
    nil : [] --> b 
    prod : [b] --> c 
    sum : [b] --> c 
    xap : [c -> a -> c * c] --> a -> c 
    yap : [a -> c * a] --> c 

  Rules:

    fold(/\x./\y.yap(xap(f, x), y), nil, z) => z 
    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)) 
    sum(x) => fold(/\y./\z.yap(xap(add, y), z), x, 0) 
    fold(mul, x, 1) => prod(x) 
    xap(f, x) => f x 
    yap(f, x) => f x 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

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:

  Alphabet:

    0 : [] --> c 
    1 : [] --> c 
    add : [c] --> a -> c 
    cons : [a * b] --> b 
    fold : [c -> a -> c * b * c] --> c 
    mul : [] --> c -> a -> c 
    nil : [] --> b 
    prod : [b] --> c 
    sum : [b] --> c 
    yap : [a -> c * a] --> c 

  Rules:

    fold(/\x./\y.yap(F(x), y), nil, X) => X 
    fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) => fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
    sum(X) => fold(/\x./\y.yap(add(x), y), X, 0) 
    fold(mul, X, 1) => prod(X) 
    yap(F, X) => F X 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  fold(/\x./\y.yap(F(x), y), nil, X) >? X 
  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) >? fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
  sum(X) >? fold(/\x./\y.yap(add(x), y), X, 0) 
  fold(mul, X, 1) >? prod(X) 
  yap(F, X) >? F X 

We use a recursive path ordering as defined in [Kop12, Chapter 5].

Argument functions:

  [[0]] = _|_ 
  [[fold(x_1, x_2, x_3)]] = fold(x_2, x_3, x_1) 
  [[prod(x_1)]] = x_1 

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}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  fold(/\x./\y.yap(F(x), y), nil, X) >= X 
  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) > fold(/\x./\y.yap(F(x), y), Y, yap(F(Z), X)) 
  sum(X) >= fold(/\x./\y.yap(add(x), y), X, _|_) 
  fold(mul, X, 1) > X 
  yap(F, X) >= @_{o -> o}(F, X) 

With these choices, we have:

  1] fold(/\x./\y.yap(F(x), y), nil, X) >= X  because [2], by (Star) 
  2] fold*(/\x./\y.yap(F(x), y), nil, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  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 
  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) 
  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) 
  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) 
  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) 
  9] cons(X, Y) > Y  because [10], by definition 
  10] cons*(X, Y) >= Y  because [11], by (Select) 
  11] Y >= Y  by (Meta) 
  12] fold*(/\x./\y.yap(F(x), y), cons(X, Y), Z) >= /\x./\y.yap(F(x), y)  because [13], by (Select) 
  13] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z)  because [14], by (Abs) 
  14] /\z.yap(F(y), z) >= /\z.yap(F(y), z)  because [15], by (Abs) 
  15] yap(F(y), x) >= yap(F(y), x)  because yap in Mul, [16] and [18], by (Fun) 
  16] F(y) >= F(y)  because [17], by (Meta) 
  17] y >= y  by (Var) 
  18] x >= x  by (Var) 
  19] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= Y  because [20], by (Select) 
  20] cons(X, Y) >= Y  because [10], by (Star) 
  21] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= yap(F(Z), X)  because fold > yap, [22] and [27], by (Copy) 
  22] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= F(Z)  because [23], by (Select) 
  23] /\z.yap(F(fold*(/\u./\v.yap(F(u), v), cons(X, Y), Z)), z) >= F(Z)  because [24], by (Eta)[Kop13:2] 
  24] F(fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z)) >= F(Z)  because [25], by (Meta) 
  25] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= Z  because [26], by (Select) 
  26] Z >= Z  by (Meta) 
  27] fold*(/\z./\u.yap(F(z), u), cons(X, Y), Z) >= X  because [28], by (Select) 
  28] cons(X, Y) >= X  because [29], by (Star) 
  29] cons*(X, Y) >= X  because [30], by (Select) 
  30] X >= X  by (Meta) 

  31] sum(X) >= fold(/\x./\y.yap(add(x), y), X, _|_)  because [32], by (Star) 
  32] sum*(X) >= fold(/\x./\y.yap(add(x), y), X, _|_)  because sum > fold, [33], [41] and [43], by (Copy) 
  33] sum*(X) >= /\y./\z.yap(add(y), z)  because [34], by (F-Abs) 
  34] sum*(X, x) >= /\z.yap(add(x), z)  because [35], by (F-Abs) 
  35] sum*(X, x, y) >= yap(add(x), y)  because sum > yap, [36] and [39], by (Copy) 
  36] sum*(X, x, y) >= add(x)  because sum > add and [37], by (Copy) 
  37] sum*(X, x, y) >= x  because [38], by (Select) 
  38] x >= x  by (Var) 
  39] sum*(X, x, y) >= y  because [40], by (Select) 
  40] y >= y  by (Var) 
  41] sum*(X) >= X  because [42], by (Select) 
  42] X >= X  by (Meta) 
  43] sum*(X) >= _|_  by (Bot) 

  44] fold(mul, X, 1) > X  because [45], by definition 
  45] fold*(mul, X, 1) >= X  because [46], by (Select) 
  46] X >= X  by (Meta) 

  47] yap(F, X) >= @_{o -> o}(F, X)  because [48], by (Star) 
  48] yap*(F, X) >= @_{o -> o}(F, X)  because yap > @_{o -> o}, [49] and [51], by (Copy) 
  49] yap*(F, X) >= F  because [50], by (Select) 
  50] F >= F  by (Meta) 
  51] yap*(F, X) >= X  because [52], by (Select) 
  52] X >= X  by (Meta) 

We can thus remove the following rules:

  fold(/\x./\y.yap(F(x), y), cons(X, Y), Z) => fold(/\z./\u.yap(F(z), u), Y, yap(F(Z), X)) 
  fold(mul, X, 1) => prod(X) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  fold(/\x./\y.yap(F(x), y), nil, X) >? X 
  sum(X) >? fold(/\x./\y.yap(add(x), y), X, 0) 
  yap(F, X) >? F X 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  0 = 0 
  add = \y0y1.y0 
  fold = \G0y1y2.y1 + y2 + G0(0,0) 
  nil = 3 
  sum = \y0.3 + 3y0 
  yap = \G0y1.y1 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[fold(/\x./\y.yap(_F0(x), y), nil, _x1)]] = 3 + x1 + F0(0,0) > x1 = [[_x1]] 
  [[sum(_x0)]] = 3 + 3x0 > x0 = [[fold(/\x./\y.yap(add(x), y), _x0, 0)]] 
  [[yap(_F0, _x1)]] = x1 + F0(x1) >= x1 + F0(x1) = [[_F0 _x1]] 

We can thus remove the following rules:

  fold(/\x./\y.yap(F(x), y), nil, X) => X 
  sum(X) => fold(/\x./\y.yap(add(x), y), X, 0) 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  yap(F, X) >? F X 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  yap = \G0y1.1 + y1 + G0(y1) 

Using this interpretation, the requirements translate to:

  [[yap(_F0, _x1)]] = 1 + x1 + F0(x1) > x1 + F0(x1) = [[_F0 _x1]] 

We can thus remove the following rules:

  yap(F, X) => F X 

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.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13:2]  C. Kop.  StarHorpo with an Eta-Rule.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/etahorpo.pdf, 2013.