We consider the system h12.

  Alphabet:

    0 : [] --> a 
    rec : [] --> (a -> b -> b) -> b -> a -> b 
    s : [] --> a -> a 

  Rules:

    rec (/\x./\y.f x y) z 0 => z 
    rec (/\x./\y.f x y) z (s u) => f (s u) (rec (/\v./\w.f v w) z u) 

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 : [] --> a 
    rec : [a -> b -> b * b * a] --> b 
    s : [a] --> a 
    ~AP1 : [a -> b -> b * a] --> b -> b 

  Rules:

    rec(/\x./\y.~AP1(F, x) y, X, 0) => X 
    rec(/\x./\y.~AP1(F, x) y, X, s(Y)) => ~AP1(F, s(Y)) rec(/\z./\u.~AP1(F, z) u, X, Y) 
    ~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:

    0 : [] --> a 
    rec : [a -> b -> b * b * a] --> b 
    s : [a] --> a 

  Rules:

    rec(/\x./\y.X(x, y), Y, 0) => Y 
    rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 

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

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

  rec(/\x./\y.X(x, y), Y, 0) >? Y 
  rec(/\x./\y.X(x, y), Y, s(Z)) >? X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 

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

Argument functions:

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

We choose Lex = {rec} and Mul = {0, s}, and the following precedence: 0 > rec > s

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

  rec(/\x./\y.X(x, y), Y, 0) >= Y 
  rec(/\x./\y.X(x, y), Y, s(Z)) > X(s(Z), rec(/\x./\y.X(x, y), Y, Z)) 

With these choices, we have:

  1] rec(/\x./\y.X(x, y), Y, 0) >= Y  because [2], by (Star) 
  2] rec*(/\x./\y.X(x, y), Y, 0) >= Y  because [3], by (Select) 
  3] Y >= Y  by (Meta) 

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

We can thus remove the following rules:

  rec(/\x./\y.X(x, y), Y, s(Z)) => X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 

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

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

  rec(/\x./\y.X(x, y), Y, 0) >? Y 

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

We choose Lex = {} and Mul = {0, rec}, and the following precedence: 0 > rec

With these choices, we have:

  1] rec(/\x./\y.X(x, y), Y, 0) > Y  because [2], by definition 
  2] rec*(/\x./\y.X(x, y), Y, 0) >= Y  because [3], by (Select) 
  3] Y >= Y  by (Meta) 

We can thus remove the following rules:

  rec(/\x./\y.X(x, y), Y, 0) => Y 

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.