We consider the system AotoYamada_05__026.

  Alphabet:

    comp : [c -> c * c -> c] --> c -> c 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 
    twice : [c -> c] --> c -> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    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 and accessible arguments in [Kop13]).

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

  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  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, static, formative):

  Dependency Pairs P_0:

    0] map#(F, cons(X, Y)) =#> map#(F, Y)   
    1] twice#(F, X) =#> comp#(F, F, X)   

  Rules R_0:

    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    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, 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 ::=

  map(F, cons(X, Y)) => cons(F X, map(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:

  map#(F, cons(X, Y)) >? map#(F, Y) 
  twice#(F, X) >? comp#(F, F, X) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

  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-twice## = \y0.3 + y0 
  comp# = \G0G1y2.0 
  cons = \y0y1.3 
  map = \G0y1.3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) 
  map# = \G0y1.0 
  twice# = \G0y1.0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[map#(_F0, _x2)]] 
  [[#argfun-twice##(comp#(_F0, _F0, _x1))]] = 3 > 0 = [[comp#(_F0, _F0, _x1)]] 
  [[map(_F0, cons(_x1, _x2))]] = 9 + 2F0(0) + 11F0(3) >= 3 = [[cons(_F0 _x1, map(_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:

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

  map#(F, cons(X, Y)) >? map#(F, Y) 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y0 + y1 
  map = \G0y1.y1 + G0(y1) + y1G0(y1) 
  map# = \G0y1.y1 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 3F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) >= 2 + x2 + F0(x1) + F0(x2) + x2F0(x2) = [[cons(_F0 _x1, map(_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.