We consider the system h02.

  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 

  Rules:

    fold (/\x./\y.f x y) nil z => z 
    fold (/\x./\y.f x y) (cons z u) v => fold (/\w./\x'.f w x') u (f v z) 
    sum x => fold (/\y./\z.add y z) x 0 
    fold (/\x./\y.mul x y) z 1 => prod z 

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.

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:

  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 
    ~AP1 : [c -> a -> c * c] --> a -> c 

  Rules:

    fold(/\x./\y.~AP1(F, x) y, nil, X) => X 
    fold(/\x./\y.~AP1(F, x) y, cons(X, Y), Z) => fold(/\z./\u.~AP1(F, z) u, Y, ~AP1(F, Z) X) 
    sum(X) => fold(/\x./\y.~AP1(add, x) y, X, 0) 
    fold(/\x./\y.mul(x, y), X, 1) => prod(X) 
    fold(/\x./\y.mul(x, y), nil, X) => X 
    fold(/\x./\y.mul(x, y), cons(X, Y), Z) => fold(/\z./\u.mul(z, u), Y, mul(Z, X)) 
    ~AP1(F, X) => F X 

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.  Additionally, we can remove some (now-)redundant rules.  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 

  Rules:

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

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

Argument functions:

  [[0]] = _|_ 
  [[prod(x_1)]] = x_1 

We choose Lex = {fold} and Mul = {1, add, cons, mul, nil, sum}, and the following precedence: 1 > cons > mul > nil > sum > add > fold

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

  fold(/\x./\y.X(x, y), nil, Y) >= Y 
  fold(/\x./\y.X(x, y), cons(Y, Z), U) > fold(/\x./\y.X(x, y), Z, X(U, Y)) 
  sum(X) >= fold(/\x./\y.add(x, y), X, _|_) 
  fold(/\x./\y.mul(x, y), X, 1) >= X 

With these choices, we have:

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

  4] fold(/\x./\y.X(x, y), cons(Y, Z), U) > fold(/\x./\y.X(x, y), Z, X(U, Y))  because [5], by definition 
  5] fold*(/\x./\y.X(x, y), cons(Y, Z), U) >= fold(/\x./\y.X(x, y), Z, X(U, Y))  because [6], [11], [14], [15] and [17], by (Stat) 
  6] /\x./\z.X(x, z) >= /\x./\z.X(x, z)  because [7], by (Abs) 
  7] /\z.X(y, z) >= /\z.X(y, z)  because [8], by (Abs) 
  8] X(y, x) >= X(y, x)  because [9] and [10], by (Meta) 
  9] y >= y  by (Var) 
  10] x >= x  by (Var) 
  11] cons(Y, Z) > Z  because [12], by definition 
  12] cons*(Y, Z) >= Z  because [13], by (Select) 
  13] Z >= Z  by (Meta) 
  14] fold*(/\z./\u.X(z, u), cons(Y, Z), U) >= /\z./\u.X(z, u)  because [6], by (Select) 
  15] fold*(/\z./\u.X(z, u), cons(Y, Z), U) >= Z  because [16], by (Select) 
  16] cons(Y, Z) >= Z  because [12], by (Star) 
  17] fold*(/\z./\u.X(z, u), cons(Y, Z), U) >= X(U, Y)  because [18], by (Select) 
  18] X(fold*(/\z./\u.X(z, u), cons(Y, Z), U), fold*(/\v./\w.X(v, w), cons(Y, Z), U)) >= X(U, Y)  because [19] and [21], by (Meta) 
  19] fold*(/\z./\u.X(z, u), cons(Y, Z), U) >= U  because [20], by (Select) 
  20] U >= U  by (Meta) 
  21] fold*(/\z./\u.X(z, u), cons(Y, Z), U) >= Y  because [22], by (Select) 
  22] cons(Y, Z) >= Y  because [23], by (Star) 
  23] cons*(Y, Z) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 

  25] sum(X) >= fold(/\x./\y.add(x, y), X, _|_)  because [26], by (Star) 
  26] sum*(X) >= fold(/\x./\y.add(x, y), X, _|_)  because sum > fold, [27], [34] and [36], by (Copy) 
  27] sum*(X) >= /\y./\z.add(y, z)  because [28], by (F-Abs) 
  28] sum*(X, x) >= /\z.add(x, z)  because [29], by (F-Abs) 
  29] sum*(X, x, y) >= add(x, y)  because sum > add, [30] and [32], by (Copy) 
  30] sum*(X, x, y) >= x  because [31], by (Select) 
  31] x >= x  by (Var) 
  32] sum*(X, x, y) >= y  because [33], by (Select) 
  33] y >= y  by (Var) 
  34] sum*(X) >= X  because [35], by (Select) 
  35] X >= X  by (Meta) 
  36] sum*(X) >= _|_  by (Bot) 

  37] fold(/\x./\y.mul(x, y), X, 1) >= X  because [38], by (Star) 
  38] fold*(/\x./\y.mul(x, y), X, 1) >= X  because [39], by (Select) 
  39] X >= X  by (Meta) 

We can thus remove the following rules:

  fold(/\x./\y.X(x, y), cons(Y, Z), U) => fold(/\z./\u.X(z, u), Z, X(U, Y)) 

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.X(x, y), nil, Y) >? Y 
  sum(X) >? fold(/\x./\y.add(x, y), X, 0) 
  fold(/\x./\y.mul(x, y), X, 1) >? prod(X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  1 = 3 
  add = \y0y1.y0 + y1 
  fold = \G0y1y2.y1 + y2 + G0(0,0) 
  mul = \y0y1.3 + 3y0 + 3y1 
  nil = 3 
  prod = \y0.y0 
  sum = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[fold(/\x./\y._x0(x, y), nil, _x1)]] = 3 + x1 + F0(0,0) > x1 = [[_x1]] 
  [[sum(_x0)]] = 3 + 3x0 > x0 = [[fold(/\x./\y.add(x, y), _x0, 0)]] 
  [[fold(/\x./\y.mul(x, y), _x0, 1)]] = 6 + x0 > x0 = [[prod(_x0)]] 

We can thus remove the following rules:

  fold(/\x./\y.X(x, y), nil, Y) => Y 
  sum(X) => fold(/\x./\y.add(x, y), X, 0) 
  fold(/\x./\y.mul(x, y), X, 1) => prod(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 +++

[Kop11]  C. Kop.  Simplifying Algebraic Functional Systems.  In Proceedings of CAI 2011, volume 6742 of LNCS.  201--215, Springer, 2011.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.