We consider the system AotoYamada_05__028.

  Alphabet:

    cons : [a * c] --> c 
    consif : [b * a * c] --> c 
    false : [] --> b 
    filter : [a -> b * c] --> c 
    nil : [] --> c 
    true : [] --> b 

  Rules:

    consif(true, x, y) => cons(x, y) 
    consif(false, x, y) => y 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) 

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] filter#(F, cons(X, Y)) =#> consif#(F X, X, filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    consif(true, X, Y) => cons(X, Y) 
    consif(false, X, Y) => Y 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 

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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 : 0, 1 

This graph has the following strongly connected components:

  P_1:

    filter#(F, cons(X, Y)) =#> filter#(F, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.

We consider the dependency pair problem (P_1, 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:

  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  consif(true, X, Y) >= cons(X, Y) 
  consif(false, X, Y) >= Y 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= consif(F X, X, filter(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y0 + 2y1 
  consif = \y0y1y2.y1 + 2y0 + 2y2 
  false = 0 
  filter = \G0y1.y1 + y1G0(y1) 
  filter# = \G0y1.2y1 + y1G0(y1) 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 4x2 + 2x2F0(2 + x1 + 2x2) + 2F0(2 + x1 + 2x2) + x1F0(2 + x1 + 2x2) > 2x2 + x2F0(x2) = [[filter#(_F0, _x2)]] 
  [[consif(true, _x0, _x1)]] = 6 + x0 + 2x1 >= 2 + x0 + 2x1 = [[cons(_x0, _x1)]] 
  [[consif(false, _x0, _x1)]] = x0 + 2x1 >= x1 = [[_x1]] 
  [[filter(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + 2x2 + 2x2F0(2 + x1 + 2x2) + 2F0(2 + x1 + 2x2) + x1F0(2 + x1 + 2x2) >= x1 + 2x2 + 2x2F0(x2) + 2F0(x1) = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_1, 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.