We consider the system kop12thesis_ex7.23.

  Alphabet:

    0 : [] --> o 
    either : [o * o] --> o 
    f : [o -> o * o * o] --> o 
    g : [o * o] --> o 
    s : [o] --> o 

  Rules:

    f(h, x, 0) => 0 
    f(h, x, s(y)) => g(y, either(y, h x)) 
    g(x, y) => f(/\z.s(0), y, 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] f#(F, X, s(Y)) =#> g#(Y, either(Y, F X))   
    1] g#(X, Y) =#> f#(/\x.s(0), Y, X)   

  Rules R_0:

    f(F, X, 0) => 0 
    f(F, X, s(Y)) => g(Y, either(Y, F X)) 
    g(X, Y) => f(/\x.s(0), Y, 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).

This combination (P_0, R_0) has no formative rules!  We will name the empty set of rules:R_1.

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:

  f#(F, X, s(Y)) >? g#(Y, either(Y, F X)) 
  g#(X, Y) >? f#(/\x.s(0), Y, X) 

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

  pi( g#(X, Y) ) = #argfun-g##(f#(/\x.s(0), Y, X)) 

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

The following interpretation satisfies the requirements:

  #argfun-g## = \y0.1 + y0 
  0 = 0 
  either = \y0y1.0 
  f# = \G0y1y2.y2 
  g# = \y0y1.0 
  s = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[f#(_F0, _x1, s(_x2))]] = 3 + 2x2 > 1 + x2 = [[#argfun-g##(f#(/\x.s(0), either(_x2, _F0 _x1), _x2))]] 
  [[#argfun-g##(f#(/\x.s(0), _x0, _x1))]] = 1 + x1 > x1 = [[f#(/\x.s(0), _x0, _x1)]] 

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.