We consider the system kop12thesis_ex2.11.

  Alphabet:

    cons : [nat * list] --> list 
    emap : [nat -> nat * list] --> list 
    nil : [] --> list 
    twice : [nat -> nat] --> nat -> nat 

  Rules:

    emap(f, nil) => nil 
    emap(f, cons(x, y)) => cons(f x, emap(/\z.twice(f) z, y)) 
    twice(f) x => f (f 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]).

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

  Dependency Pairs P_0:

    0] emap#(F, cons(X, Y)) =#> emap#(/\x.twice(F, x), Y)   
    1] emap#(F, cons(X, Y)) =#> twice#(F, Z)   

  Rules R_0:

    emap(F, nil) => nil 
    emap(F, cons(X, Y)) => cons(F X, emap(/\x.twice(F, x), Y)) 
    twice(F, X) => F (F 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).

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

  emap(F, cons(X, Y)) => cons(F X, emap(/\x.twice(F, x), 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:

  emap#(F, cons(X, Y)) >? emap#(/\x.twice(F, x), Y) 
  emap#(F, cons(X, Y)) >? twice#(F, Z) 
  emap(F, cons(X, Y)) >= cons(F X, emap(/\x.twice(F, x), Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y1 
  emap = \G0y1.3y1 + 2y1G0(y1) + 2G0(y1) 
  emap# = \G0y1.3 + y1 
  twice = \G0y1.0 
  twice# = \G0y1.0 

Using this interpretation, the requirements translate to:

  [[emap#(_F0, cons(_x1, _x2))]] = 6 + 2x2 > 3 + x2 = [[emap#(/\x.twice(_F0, x), _x2)]] 
  [[emap#(_F0, cons(_x1, _x2))]] = 6 + 2x2 > 0 = [[twice#(_F0, _x3)]] 
  [[emap(_F0, cons(_x1, _x2))]] = 9 + 6x2 + 4x2F0(3 + 2x2) + 8F0(3 + 2x2) >= 3 + 6x2 = [[cons(_F0 _x1, emap(/\x.twice(_F0, x), _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, 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.