We consider the system AotoYamada_05__005.

  Alphabet:

    0 : [] --> a 
    add : [] --> a -> a -> a 
    curry : [a -> a -> a] --> a -> a -> a 
    plus : [] --> a -> a -> a 
    s : [a] --> a 

  Rules:

    plus 0 x => x 
    plus s(x) y => s(plus x y) 
    curry(f) x y => f x y 
    add => curry(plus) 

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:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  curry(F, X, Y) => F X Y 
  add(X, Y) => curry(/\x./\y.plus(x, y), X, Y) 

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

  Dependency Pairs P_0:

    0] plus#(s(X), Y) =#> plus#(X, Y)   
    1] add#(X, Y) =#> curry#(/\x./\y.plus(x, y), X, Y)   
    2] add#(X, Y) =#> plus#(Z, U)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    curry(F, X, Y) => F X Y 
    add(X, Y) => curry(/\x./\y.plus(x, y), X, Y) 

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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 : 0 
    * 1 :  
    * 2 : 0 

This graph has the following strongly connected components:

  P_1:

    plus#(s(X), Y) =#> plus#(X, Y)   

By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).

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:

  plus#(s(X), Y) >? plus#(X, Y) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  curry(F, X, Y) >= F X Y 
  add(X, Y) >= curry(/\x./\y.plus(x, y), X, Y) 

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

  pi( curry(F, X, Y) ) = #argfun-curry#(F X Y) 
  pi( add(X, Y) ) = #argfun-add#(#argfun-curry#((/\x./\y.plus(x, y)) X Y)) 

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

The following interpretation satisfies the requirements:

  #argfun-add# = \y0.3 + y0 
  #argfun-curry# = \y0.3 + y0 
  0 = 0 
  add = \y0y1.0 
  curry = \G0y1y2.0 
  plus = \y0y1.y1 + 2y0 
  plus# = \y0y1.y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 1 + x0 > x0 = [[plus#(_x0, _x1)]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 2 + x1 + 2x0 >= 1 + x1 + 2x0 = [[s(plus(_x0, _x1))]] 
  [[#argfun-curry#(_F0 _x1 _x2)]] = 3 + F0(x1,x2) >= F0(x1,x2) = [[_F0 _x1 _x2]] 
  [[#argfun-add#(#argfun-curry#((/\x./\y.plus(x, y)) _x0 _x1))]] = 6 + x1 + 2x0 >= 3 + x1 + 2x0 = [[#argfun-curry#((/\x./\y.plus(x, y)) _x0 _x1)]] 

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.