We consider the system AotoYamada_05__019.

  Alphabet:

    comp : [a -> a * a -> a] --> a -> a 
    twice : [a -> a] --> a -> a 

  Rules:

    comp(f, g) x => f (g x) 
    twice(f) => comp(f, f) 

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 in [Kop12, Ch. 7.8]).

In order to do so, we start by eta-expanding the system, which gives:

  comp(F, G, X) => F (G X) 
  twice(F, X) => comp(F, F, X) 

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

  Dependency Pairs P_0:

    0] twice#(F, X) =#> comp#(F, F, X)   

  Rules R_0:

    comp(F, G, X) => F (G X) 
    twice(F, X) => comp(F, F, X) 

Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.

We consider the dependency pair problem (P_0, R_0, minimal, 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:

  twice#(F, X) >? comp#(F, F, X) 
  comp(F, G, X) >= F (G X) 
  twice(F, X) >= comp(F, F, X) 

We apply [Kop12, Thm. 6.75] and use the following argument functions:

  pi( comp(F, G, X) ) = #argfun-comp#(F (G X)) 
  pi( twice(F, X) ) = #argfun-twice#(#argfun-comp#(F (F X))) 
  pi( twice#(F, X) ) = #argfun-twice##(comp#(F, F, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-comp# = \y0.3 + y0 
  #argfun-twice# = \y0.3 + y0 
  #argfun-twice## = \y0.1 + y0 
  comp = \G0G1y2.0 
  comp# = \G0G1y2.0 
  twice = \G0y1.0 
  twice# = \G0y1.0 

Using this interpretation, the requirements translate to:

  [[#argfun-twice##(comp#(_F0, _F0, _x1))]] = 1 > 0 = [[comp#(_F0, _F0, _x1)]] 
  [[#argfun-comp#(_F0 (_F1 _x2))]] = 3 + F0(F1(x2)) >= F0(F1(x2)) = [[_F0 (_F1 _x2)]] 
  [[#argfun-twice#(#argfun-comp#(_F0 (_F0 _x1)))]] = 6 + F0(F0(x1)) >= 3 + F0(F0(x1)) = [[#argfun-comp#(_F0 (_F0 _x1))]] 

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