We consider the system 431.

  Alphabet:

    0 : [] --> nat 
    rec : [nat -> A -> A * A * nat] --> A 
    s : [nat] --> nat 

  Rules:

    rec(/\x./\y.X(x, y), Y, 0) => Y 
    rec(/\x./\y.X(x, y), Y, s(Z)) => X(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(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(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(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 + y2 + G0(0,0) + 2y2y2G0(y2,y2) + 2G0(y2,y2) + 3y1y2G0(y1,y2) + 3y1y2G0(y2,y1) 
  s = \y0.3 + 3y0 
  ~c0 = 0 
  ~c1 = 0 

Using this interpretation, the requirements translate to:

  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 6 + 3x2 + F0(0,0) + 9x1x2F0(x1,3 + 3x2) + 9x1x2F0(3 + 3x2,x1) + 9x1F0(x1,3 + 3x2) + 9x1F0(3 + 3x2,x1) + 18x2x2F0(3 + 3x2,3 + 3x2) + 20F0(3 + 3x2,3 + 3x2) + 36x2F0(3 + 3x2,3 + 3x2) > F0(x2,0) = [[_x0(_x2, rec(/\x./\y._x0(x, y), _x1, _x2))]] 
  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 6 + 3x2 + F0(0,0) + 9x1x2F0(x1,3 + 3x2) + 9x1x2F0(3 + 3x2,x1) + 9x1F0(x1,3 + 3x2) + 9x1F0(3 + 3x2,x1) + 18x2x2F0(3 + 3x2,3 + 3x2) + 20F0(3 + 3x2,3 + 3x2) + 36x2F0(3 + 3x2,3 + 3x2) > 3 + x2 + F0(0,0) + 2x2x2F0(x2,x2) + 2F0(x2,x2) + 3x1x2F0(x1,x2) + 3x1x2F0(x2,x1) = [[rec#(/\x./\y._x0(x, y), _x1, _x2)]] 
  [[rec#(/\x./\y._x0(x, y), _x1, s(_x2))]] = 6 + 3x2 + F0(0,0) + 9x1x2F0(x1,3 + 3x2) + 9x1x2F0(3 + 3x2,x1) + 9x1F0(x1,3 + 3x2) + 9x1F0(3 + 3x2,x1) + 18x2x2F0(3 + 3x2,3 + 3x2) + 20F0(3 + 3x2,3 + 3x2) + 36x2F0(3 + 3x2,3 + 3x2) > F0(0,0) = [[_x0(~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 +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.