We consider the system h51.

  Alphabet:

    cons : [] --> a -> alist -> alist 
    foldl : [] --> (a -> a -> a) -> a -> alist -> a 
    nil : [] --> alist 

  Rules:

    foldl (/\x./\y.f x y) z nil => z 
    foldl (/\x./\y.f x y) z (cons u v) => foldl (/\w./\x'.f w x') (f z u) v 

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:

    cons : [a * alist] --> alist 
    foldl : [a -> a -> a * a * alist] --> a 
    nil : [] --> alist 
    ~AP1 : [a -> a -> a * a] --> a -> a 

  Rules:

    foldl(/\x./\y.~AP1(F, x) y, X, nil) => X 
    foldl(/\x./\y.~AP1(F, x) y, X, cons(Y, Z)) => foldl(/\z./\u.~AP1(F, z) u, ~AP1(F, X) Y, Z) 
    ~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.  This gives:

  Alphabet:

    cons : [a * alist] --> alist 
    foldl : [a -> a -> a * a * alist] --> a 
    nil : [] --> alist 

  Rules:

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

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

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

  foldl(/\x./\y.X(x, y), Y, nil) >? Y 
  foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >? foldl(/\z./\u.X(z, u), X(Y, Z), U) 

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

Argument functions:

  [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_2, x_1) 

We choose Lex = {foldl} and Mul = {cons, nil}, and the following precedence: cons > foldl > nil

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

  foldl(/\x./\y.X(x, y), Y, nil) > Y 
  foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U) 

With these choices, we have:

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

  4] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U)  because [5], by (Star) 
  5] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U)  because [6], [9], [16] and [24], by (Stat) 
  6] cons(Z, U) > U  because [7], by definition 
  7] cons*(Z, U) >= U  because [8], by (Select) 
  8] U >= U  by (Meta) 
  9] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y)  because [10], by (F-Abs) 
  10] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= /\x.X(z, x)  because [11], by (Select) 
  11] /\x.X(foldl*(/\y./\v.X(y, v), Y, cons(Z, U), z), x) >= /\x.X(z, x)  because [12], by (Abs) 
  12] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z), u) >= X(z, u)  because [13] and [15], by (Meta) 
  13] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= z  because [14], by (Select) 
  14] z >= z  by (Var) 
  15] u >= u  by (Var) 
  16] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Y, Z)  because [17], by (Select) 
  17] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z)  because [18] and [20], by (Meta) 
  18] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Y  because [19], by (Select) 
  19] Y >= Y  by (Meta) 
  20] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [21], by (Select) 
  21] cons(Z, U) >= Z  because [22], by (Star) 
  22] cons*(Z, U) >= Z  because [23], by (Select) 
  23] Z >= Z  by (Meta) 
  24] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= U  because [25], by (Select) 
  25] cons(Z, U) >= U  because [7], by (Star) 

We can thus remove the following rules:

  foldl(/\x./\y.X(x, y), Y, nil) => 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]):

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

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

Argument functions:

  [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_2, x_1) 

We choose Lex = {foldl} and Mul = {cons}, and the following precedence: cons > foldl

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

  foldl(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl(/\x./\y.X(x, y), X(Y, Z), U) 

With these choices, we have:

  1] foldl(/\x./\y.X(x, y), Y, cons(Z, U)) > foldl(/\x./\y.X(x, y), X(Y, Z), U)  because [2], by definition 
  2] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= foldl(/\x./\y.X(x, y), X(Y, Z), U)  because [3], [6], [14] and [22], by (Stat) 
  3] cons(Z, U) > U  because [4], by definition 
  4] cons*(Z, U) >= U  because [5], by (Select) 
  5] U >= U  by (Meta) 
  6] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= /\x./\y.X(x, y)  because [7], by (F-Abs) 
  7] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z) >= /\x.X(z, x)  because [8], by (F-Abs) 
  8] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= X(z, u)  because [9], by (Select) 
  9] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u), foldl*(/\v./\w.X(v, w), Y, cons(Z, U), z, u)) >= X(z, u)  because [10] and [12], by (Meta) 
  10] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= z  because [11], by (Select) 
  11] z >= z  by (Var) 
  12] foldl*(/\x./\y.X(x, y), Y, cons(Z, U), z, u) >= u  because [13], by (Select) 
  13] u >= u  by (Var) 
  14] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= X(Y, Z)  because [15], by (Select) 
  15] X(foldl*(/\x./\y.X(x, y), Y, cons(Z, U)), foldl*(/\v./\w.X(v, w), Y, cons(Z, U))) >= X(Y, Z)  because [16] and [18], by (Meta) 
  16] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Y  because [17], by (Select) 
  17] Y >= Y  by (Meta) 
  18] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= Z  because [19], by (Select) 
  19] cons(Z, U) >= Z  because [20], by (Star) 
  20] cons*(Z, U) >= Z  because [21], by (Select) 
  21] Z >= Z  by (Meta) 
  22] foldl*(/\x./\y.X(x, y), Y, cons(Z, U)) >= U  because [23], by (Select) 
  23] cons(Z, U) >= U  because [4], by (Star) 

We can thus remove the following rules:

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

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.