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).

The formative rules of (P_0, R_0) are R_1 ::=

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

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, static, formative) by (P_0, R_1, static, formative).

Thus, the original system is terminating if (P_0, R_1, static, formative) is finite.

We consider the dependency pair problem (P_0, R_1, 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)) >? consif#(F X, X, filter(F, Y)) 
  filter#(F, cons(X, Y)) >? filter#(F, Y) 
  consif(true, X, Y) >= cons(X, Y) 
  consif(false, X, Y) >= Y 
  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 = \y0y1.0 
  consif = \y0y1y2.2y2 
  consif# = \y0y1y2.0 
  false = 0 
  filter = \G0y1.0 
  filter# = \G0y1.3 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[consif#(_F0 _x1, _x1, filter(_F0, _x2))]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[filter#(_F0, _x2)]] 
  [[consif(true, _x0, _x1)]] = 2x1 >= 0 = [[cons(_x0, _x1)]] 
  [[consif(false, _x0, _x1)]] = 2x1 >= x1 = [[_x1]] 
  [[filter(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, static, formative) by (P_1, R_1, static, formative), where P_1 consists of:

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

Thus, the original system is terminating if (P_1, R_1, static, formative) is finite.

We consider the dependency pair problem (P_1, R_1, 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, 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 + y1 + 2y0 
  consif = \y0y1y2.y0 + y2 + 2y1 
  false = 0 
  filter = \G0y1.2y1 + 2y1G0(y1) 
  filter# = \G0y1.y1 
  true = 3 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 2 + x2 + 2x1 > x2 = [[filter#(_F0, _x2)]] 
  [[consif(true, _x0, _x1)]] = 3 + x1 + 2x0 >= 2 + x1 + 2x0 = [[cons(_x0, _x1)]] 
  [[consif(false, _x0, _x1)]] = x1 + 2x0 >= x1 = [[_x1]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) >= 2x1 + 2x2 + F0(x1) + 2x2F0(x2) = [[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_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.
[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.