We consider the system loopy.

  Alphabet:

    f : [N -> N * N] --> N 
    g : [] --> N -> N 
    h : [N] --> N 

  Rules:

    f(g, x) => h(x) 
    f(i, x) => i x 
    h(x) => f(/\y.y, x) 

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]).

In order to do so, we start by eta-expanding the system, which gives:

  f(/\x.g(x), X) => h(X) 
  f(F, X) => F X 
  h(X) => f(/\x.x, X) 

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

  Dependency Pairs P_0:

    0] f#(/\x.g(x), X) =#> h#(X)   
    1] h#(X) =#> f#(/\x.x, X)   

  Rules R_0:

    f(/\x.g(x), X) => h(X) 
    f(F, X) => F X 
    h(X) => f(/\x.x, X) 

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:

  f#(/\x.g(x), X) >? h#(X) 
  h#(X) >? f#(/\x.x, X) 
  f(/\x.g(x), X) >= h(X) 
  f(F, X) >= F X 
  h(X) >= f(/\x.x, X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( h(X) ) = #argfun-h#(f(/\x.x, X)) 
  pi( h#(X) ) = #argfun-h##(f#(/\x.x, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-h# = \y0.y0 
  #argfun-h## = \y0.1 + y0 
  f = \G0y1.2G0(y1) + y1G0(y1) 
  f# = \G0y1.G0(0) + y1G0(y1) 
  g = \y0.3 + 2y0 
  h = \y0.0 
  h# = \y0.0 

Using this interpretation, the requirements translate to:

  [[f#(/\x.g(x), _x0)]] = 3 + 2x0x0 + 3x0 > 1 + x0x0 = [[#argfun-h##(f#(/\x.x, _x0))]] 
  [[#argfun-h##(f#(/\x.x, _x0))]] = 1 + x0x0 > x0x0 = [[f#(/\x.x, _x0)]] 
  [[f(/\x.g(x), _x0)]] = 6 + 2x0x0 + 7x0 >= 2x0 + x0x0 = [[#argfun-h#(f(/\x.x, _x0))]] 
  [[f(_F0, _x1)]] = 2F0(x1) + x1F0(x1) >= F0(x1) = [[_F0 _x1]] 
  [[#argfun-h#(f(/\x.x, _x0))]] = 2x0 + x0x0 >= 2x0 + x0x0 = [[f(/\x.x, _x0)]] 

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.