We consider the system rec.

  Alphabet:

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

  Rules:

    rec(0, x, f) => x 
    rec(s(x), y, f) => f x rec(x, y, f) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

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

After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):

  Dependency Pairs P_0:

    0] rec#(s(X), Y, F) =#> F(X, rec(X, Y, F))   
    1] rec#(s(X), Y, F) =#> rec#(X, Y, F)   

  Rules R_0:

    rec(0, X, F) => X 
    rec(s(X), Y, F) => F X rec(X, Y, F) 

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, F) >? F(X, rec(X, Y, F)) 
  rec#(s(X), Y, F) >? rec#(X, Y, F) 

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 + 2y0y0G2(y0,y0) + 2G2(y0,y0) + 3y0y1G2(y0,y1) + 3y0y1G2(y1,y0) 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[rec#(s(_x0), _x1, _F2)]] = 6 + 3x0 + 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) = [[_F2(_x0, rec(_x0, _x1, _F2))]] 
  [[rec#(s(_x0), _x1, _F2)]] = 6 + 3x0 + 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 + 2x0x0F2(x0,x0) + 2F2(x0,x0) + 3x0x1F2(x0,x1) + 3x0x1F2(x1,x0) = [[rec#(_x0, _x1, _F2)]] 

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.