We consider the system Applicative_05__Ex9Maps.

  Alphabet:

    cons : [d * e] --> e 
    g : [] --> b 
    map!fac62201 : [d -> d * e] --> e 
    map!fac62202 : [d -> a -> d * a * e] --> e 
    map!fac62203 : [b -> d -> c -> d * b * c * e] --> e 

  Rules:

    map!fac62201(f, cons(x, y)) => cons(f x, map!fac62201(f, y)) 
    map!fac62202(f, x, cons(y, z)) => cons(f y x, map!fac62202(f, x, z)) 
    map!fac62203(f, g, x, cons(y, z)) => cons(f g y x, map!fac62203(f, g, x, z)) 

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] map!fac62201#(F, cons(X, Y)) =#> map!fac62201#(F, Y)   
    1] map!fac62202#(F, X, cons(Y, Z)) =#> map!fac62202#(F, X, Z)   
    2] map!fac62203#(F, g, X, cons(Y, Z)) =#> map!fac62203#(F, g, X, Z)   

  Rules R_0:

    map!fac62201(F, cons(X, Y)) => cons(F X, map!fac62201(F, Y)) 
    map!fac62202(F, X, cons(Y, Z)) => cons(F Y X, map!fac62202(F, X, Z)) 
    map!fac62203(F, g, X, cons(Y, Z)) => cons(F g Y X, map!fac62203(F, g, X, Z)) 

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:

  map!fac62201#(F, cons(X, Y)) >? map!fac62201#(F, Y) 
  map!fac62202#(F, X, cons(Y, Z)) >? map!fac62202#(F, X, Z) 
  map!fac62203#(F, g, X, cons(Y, Z)) >? map!fac62203#(F, g, X, Z) 
  map!fac62201(F, cons(X, Y)) >= cons(F X, map!fac62201(F, Y)) 
  map!fac62202(F, X, cons(Y, Z)) >= cons(F Y X, map!fac62202(F, X, Z)) 
  map!fac62203(F, g, X, cons(Y, Z)) >= cons(F g Y X, map!fac62203(F, g, X, Z)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y1 
  g = 0 
  map!fac62201 = \G0y1.y1 
  map!fac62201# = \G0y1.0 
  map!fac62202 = \G0y1y2.y2 
  map!fac62202# = \G0y1y2.y2 
  map!fac62203 = \G0y1y2y3.2 + 2y3 
  map!fac62203# = \G0y1y2y3.y3 

Using this interpretation, the requirements translate to:

  [[map!fac62201#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map!fac62201#(_F0, _x2)]] 
  [[map!fac62202#(_F0, _x1, cons(_x2, _x3))]] = 1 + x3 > x3 = [[map!fac62202#(_F0, _x1, _x3)]] 
  [[map!fac62203#(_F0, g, _x1, cons(_x2, _x3))]] = 1 + x3 > x3 = [[map!fac62203#(_F0, g, _x1, _x3)]] 
  [[map!fac62201(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map!fac62201(_F0, _x2))]] 
  [[map!fac62202(_F0, _x1, cons(_x2, _x3))]] = 1 + x3 >= 1 + x3 = [[cons(_F0 _x2 _x1, map!fac62202(_F0, _x1, _x3))]] 
  [[map!fac62203(_F0, g, _x1, cons(_x2, _x3))]] = 4 + 2x3 >= 3 + 2x3 = [[cons(_F0 g _x2 _x1, map!fac62203(_F0, g, _x1, _x3))]] 

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:

  map!fac62201#(F, cons(X, Y)) =#> map!fac62201#(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:

  map!fac62201#(F, cons(X, Y)) >? map!fac62201#(F, Y) 
  map!fac62201(F, cons(X, Y)) >= cons(F X, map!fac62201(F, Y)) 
  map!fac62202(F, X, cons(Y, Z)) >= cons(F Y X, map!fac62202(F, X, Z)) 
  map!fac62203(F, g, X, cons(Y, Z)) >= cons(F g Y X, map!fac62203(F, g, X, Z)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + 2y1 
  g = 1 
  map!fac62201 = \G0y1.y1 
  map!fac62201# = \G0y1.y1 
  map!fac62202 = \G0y1y2.y2 
  map!fac62203 = \G0y1y2y3.y3 + y1y3y3G0(y3,y3,y1) 

Using this interpretation, the requirements translate to:

  [[map!fac62201#(_F0, cons(_x1, _x2))]] = 1 + 2x2 > x2 = [[map!fac62201#(_F0, _x2)]] 
  [[map!fac62201(_F0, cons(_x1, _x2))]] = 1 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, map!fac62201(_F0, _x2))]] 
  [[map!fac62202(_F0, _x1, cons(_x2, _x3))]] = 1 + 2x3 >= 1 + 2x3 = [[cons(_F0 _x2 _x1, map!fac62202(_F0, _x1, _x3))]] 
  [[map!fac62203(_F0, g, _x1, cons(_x2, _x3))]] = 1 + 2x3 + F0(1 + 2x3,1 + 2x3,1) + 4x3x3F0(1 + 2x3,1 + 2x3,1) + 4x3F0(1 + 2x3,1 + 2x3,1) >= 1 + 2x3 + 2x3x3F0(x3,x3,1) = [[cons(_F0 g _x2 _x1, map!fac62203(_F0, g, _x1, _x3))]] 

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.