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 static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

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

  Dependency Pairs P_0:

    0] 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, static, formative) is finite.

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

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

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

Since this representation is not advantageous for the higher-order recursive path ordering, we present the strict requirements in their unextended form, which is not problematic since for any F, s and substituion gamma: (F s)gamma beta-reduces to F(s)gamma.)

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

Argument functions:

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

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

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

  s(X) > X 
  rec(0, X, F) >= X 
  rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) 

With these choices, we have:

  1] s(X) > X  because [2], by definition 
  2] s*(X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] rec(0, X, F) >= X  because [5], by (Star) 
  5] rec*(0, X, F) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 

  7] rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because [8], by (Star) 
  8] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F))  because rec > @_{o -> o}, [9] and [14], by (Copy) 
  9] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X)  because rec > @_{o -> o -> o}, [10] and [12], by (Copy) 
  10] rec*(s(X), Y, F) >= F  because [11], by (Select) 
  11] F >= F  by (Meta) 
  12] rec*(s(X), Y, F) >= X  because [13], by (Select) 
  13] s(X) >= X  because [2], by (Star) 
  14] rec*(s(X), Y, F) >= rec(X, Y, F)  because rec in Mul, [15], [16] and [17], by (Stat) 
  15] s(X) > X  because [2], by definition 
  16] Y >= Y  by (Meta) 
  17] F >= F  by (Meta) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0).  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.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.