We consider the system h59.

  Alphabet:

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

  Rules:

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

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

  Rules:

    rec(0, X, /\x./\y.~AP1(F, x) y) => X 
    rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) 
    ~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 : [] --> nat 
    rec : [nat * a * nat -> a -> a] --> a 
    s : [nat] --> nat 

  Rules:

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

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

  Rules R_0:

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

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

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

The following interpretation satisfies the requirements:

  rec = \y0y1G2.0 
  rec# = \y0y1G2.3 + y0 + G2(0,0) + 2y0y0G2(y0,y0) + 2G2(y0,y0) + 3y0y1G2(y0,y1) + 3y0y1G2(y1,y0) 
  s = \y0.3 + 3y0 
  ~c0 = 0 
  ~c1 = 0 

Using this interpretation, the requirements translate to:

  [[rec#(s(_x0), _x1, /\x./\y._x2(x, y))]] = 6 + 3x0 + F2(0,0) + 9x0x1F2(x1,3 + 3x0) + 9x0x1F2(3 + 3x0,x1) + 9x1F2(x1,3 + 3x0) + 9x1F2(3 + 3x0,x1) + 18x0x0F2(3 + 3x0,3 + 3x0) + 20F2(3 + 3x0,3 + 3x0) + 36x0F2(3 + 3x0,3 + 3x0) > F2(x0,0) = [[_x2(_x0, rec(_x0, _x1, /\x./\y._x2(x, y)))]] 
  [[rec#(s(_x0), _x1, /\x./\y._x2(x, y))]] = 6 + 3x0 + F2(0,0) + 9x0x1F2(x1,3 + 3x0) + 9x0x1F2(3 + 3x0,x1) + 9x1F2(x1,3 + 3x0) + 9x1F2(3 + 3x0,x1) + 18x0x0F2(3 + 3x0,3 + 3x0) + 20F2(3 + 3x0,3 + 3x0) + 36x0F2(3 + 3x0,3 + 3x0) > 3 + x0 + F2(0,0) + 2x0x0F2(x0,x0) + 2F2(x0,x0) + 3x0x1F2(x0,x1) + 3x0x1F2(x1,x0) = [[rec#(_x0, _x1, /\x./\y._x2(x, y))]] 
  [[rec#(s(_x0), _x1, /\x./\y._x2(x, y))]] = 6 + 3x0 + F2(0,0) + 9x0x1F2(x1,3 + 3x0) + 9x0x1F2(3 + 3x0,x1) + 9x1F2(x1,3 + 3x0) + 9x1F2(3 + 3x0,x1) + 18x0x0F2(3 + 3x0,3 + 3x0) + 20F2(3 + 3x0,3 + 3x0) + 36x0F2(3 + 3x0,3 + 3x0) > F2(0,0) = [[_x2(~c0, ~c1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1).  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 +++

[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.