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 the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.

We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] rec#(/\x./\y.X(x, y), Y, s(Z)) =#> X(s(Z), rec(/\z./\u.X(z, u), Y, Z))   
    1] rec#(/\x./\y.X(x, y), Y, s(Z)) =#> rec#(/\z./\u.X(z, u), Y, Z) {X : 2}  
    2] rec#(/\x./\y.X(x, y), Y, s(Z)) =#> X(z, u) {X : 2}  

  Rules R_0:

    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)) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, formative).

This combination (P_0, R_0) has no formative rules!  We will name the empty set of rules:R_1.

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative).

Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_1, minimal, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:

  rec#(/\x./\y.X(x, y), Y, s(Z)) >? X(s(Z), rec(/\z./\u.X(z, u), Y, Z)) 
  rec#(/\x./\y.X(x, y), Y, s(Z)) >? rec#(/\z./\u.X(z, u), Y, Z) 
  rec#(/\x./\y.X(x, y), Y, s(Z)) >? X(~c0, ~c1) 

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

The following interpretation satisfies the requirements:

  rec = \G0y1y2.0 
  rec# = \G0y1y2.3 + G0(0,0) 
  s = \y0.0 
  ~c0 = 0 
  ~c1 = 0 

Using this interpretation, the requirements translate to:

  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 3 + F0(0,0) > F0(0,0) = [[_x0(s(_x2), rec(/\x./\y._x0(x, y), _x1, _x2))]] 
  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 3 + F0(0,0) >= 3 + F0(0,0) = [[rec#(/\x./\y._x0(x, y), _x1, _x2)]] 
  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 3 + F0(0,0) > F0(0,0) = [[_x0(~c0, ~c1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of:

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

Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.

We consider the dependency pair problem (P_1, R_1, minimal, formative).

We apply the subterm criterion with the following projection function:

  nu(rec#) = 3 

Thus, we can orient the dependency pairs as follows:

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

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_1, minimal, f) by ({}, R_1, minimal, f).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[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.