We consider the system AotoYamada_05__015.

  Alphabet:

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

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), 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)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    1] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    2] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

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

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

The following interpretation satisfies the requirements:

  cons = \y0y1.0 
  false = 0 
  filter = \G0y1.2 
  filter# = \G0y1.3 
  if = \y0y1y2.y1 + y2 
  if# = \y0y1y2.0 
  nil = 0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[filter#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[if#(_F0 _x1, cons(_x1, filter(_F0, _x2)), filter(_F0, _x2))]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[filter#(_F0, _x2)]] 
  [[filter#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[filter#(_F0, _x2)]] 
  [[if(true, _x0, _x1)]] = x0 + x1 >= x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = x0 + x1 >= x1 = [[_x1]] 
  [[filter(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 2 >= 2 = [[if(_F0 _x1, cons(_x1, filter(_F0, _x2)), 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_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

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

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

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].

We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, filter#, if, nil, true}, and the following precedence: false > filter# > filter > @_{o -> o} > cons > if > nil > true

With these choices, we have:

  1] filter#(F, cons(X, Y)) >= filter#(F, Y)  because [2], by (Star) 
  2] filter#*(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [3] and [4], by (Stat) 
  3] F >= F  by (Meta) 
  4] cons(X, Y) > Y  because [5], by definition 
  5] cons*(X, Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 

  7] filter#(F, cons(X, Y)) > filter#(F, Y)  because [8], by definition 
  8] filter#*(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [3] and [4], by (Stat) 

  9] if(true, X, Y) >= X  because [10], by (Star) 
  10] if*(true, X, Y) >= X  because [11], by (Select) 
  11] X >= X  by (Meta) 

  12] if(false, X, Y) >= Y  because [13], by (Star) 
  13] if*(false, X, Y) >= Y  because [14], by (Select) 
  14] Y >= Y  by (Meta) 

  15] filter(F, nil) >= nil  because [16], by (Star) 
  16] filter*(F, nil) >= nil  because filter > nil, by (Copy) 

  17] filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because [18], by (Star) 
  18] filter*(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because filter > if, [19], [25] and [26], by (Copy) 
  19] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [20] and [21], by (Copy) 
  20] filter*(F, cons(X, Y)) >= F  because [3], by (Select) 
  21] filter*(F, cons(X, Y)) >= X  because [22], by (Select) 
  22] cons(X, Y) >= X  because [23], by (Star) 
  23] cons*(X, Y) >= X  because [24], by (Select) 
  24] X >= X  by (Meta) 
  25] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y))  because filter > cons, [21] and [26], by (Copy) 
  26] filter*(F, cons(X, Y)) >= filter(F, Y)  because filter in Mul, [3] and [4], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

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

Thus, the original system is terminating if (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, 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) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  filter(F, nil) >= nil 
  filter(F, cons(X, Y)) >= if(F X, cons(X, filter(F, Y)), filter(F, Y)) 

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:

  [[nil]] = _|_ 

We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, filter#, if, true}, and the following precedence: false > filter# > filter > @_{o -> o} > cons > if > true

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

  filter#(F, cons(X, Y)) > filter#(F, Y) 
  if(true, X, Y) >= X 
  if(false, X, Y) >= Y 
  filter(F, _|_) >= _|_ 
  filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y)) 

With these choices, we have:

  1] filter#(F, cons(X, Y)) > filter#(F, Y)  because [2], by definition 
  2] filter#*(F, cons(X, Y)) >= filter#(F, Y)  because filter# in Mul, [3] and [4], by (Stat) 
  3] F >= F  by (Meta) 
  4] cons(X, Y) > Y  because [5], by definition 
  5] cons*(X, Y) >= Y  because [6], by (Select) 
  6] Y >= Y  by (Meta) 

  7] if(true, X, Y) >= X  because [8], by (Star) 
  8] if*(true, X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 

  10] if(false, X, Y) >= Y  because [11], by (Star) 
  11] if*(false, X, Y) >= Y  because [12], by (Select) 
  12] Y >= Y  by (Meta) 

  13] filter(F, _|_) >= _|_  by (Bot) 

  14] filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because [15], by (Star) 
  15] filter*(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y))  because filter > if, [16], [22] and [23], by (Copy) 
  16] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because filter > @_{o -> o}, [17] and [18], by (Copy) 
  17] filter*(F, cons(X, Y)) >= F  because [3], by (Select) 
  18] filter*(F, cons(X, Y)) >= X  because [19], by (Select) 
  19] cons(X, Y) >= X  because [20], by (Star) 
  20] cons*(X, Y) >= X  because [21], by (Select) 
  21] X >= X  by (Meta) 
  22] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y))  because filter > cons, [18] and [23], by (Copy) 
  23] filter*(F, cons(X, Y)) >= filter(F, Y)  because filter in Mul, [3] and [4], by (Stat) 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, 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.