We consider the system Applicative_05__Ex6Folding.

  Alphabet:

    0 : [] --> c 
    1 : [] --> c 
    add : [] --> c -> a -> c 
    cons : [a * b] --> b 
    fold : [c -> a -> c * b * c] --> c 
    mul : [] --> c -> a -> c 
    nil : [] --> b 
    prod : [b] --> c 
    sum : [b] --> c 

  Rules:

    fold(f, nil, x) => x 
    fold(f, cons(x, y), z) => fold(f, y, f z x) 
    sum(x) => fold(add, x, 0) 
    fold(mul, x, 1) => prod(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]).

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

  fold(F, nil, X) => X 
  fold(F, cons(X, Y), Z) => fold(F, Y, F Z X) 
  sum(X) => fold(/\x./\y.add(x, y), X, 0) 
  fold(/\x./\y.mul(x, y), X, 1) => prod(X) 

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

  Dependency Pairs P_0:

    0] fold#(F, cons(X, Y), Z) =#> fold#(F, Y, F Z X)   
    1] sum#(X) =#> fold#(/\x./\y.add(x, y), X, 0)   

  Rules R_0:

    fold(F, nil, X) => X 
    fold(F, cons(X, Y), Z) => fold(F, Y, F Z X) 
    sum(X) => fold(/\x./\y.add(x, y), X, 0) 
    fold(/\x./\y.mul(x, y), X, 1) => prod(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:

  fold#(F, cons(X, Y), Z) >? fold#(F, Y, F Z X) 
  sum#(X) >? fold#(/\x./\y.add(x, y), X, 0) 

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

  pi( sum#(X) ) = #argfun-sum##(fold#(/\x./\y.add(x, y), X, 0)) 

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

The following interpretation satisfies the requirements:

  #argfun-sum## = \y0.3 + y0 
  0 = 0 
  add = \y0y1.0 
  cons = \y0y1.3 + 2y1 
  fold# = \G0y1y2.y1 
  sum# = \y0.0 

Using this interpretation, the requirements translate to:

  [[fold#(_F0, cons(_x1, _x2), _x3)]] = 3 + 2x2 > x2 = [[fold#(_F0, _x2, _F0 _x3 _x1)]] 
  [[#argfun-sum##(fold#(/\x./\y.add(x, y), _x0, 0))]] = 3 + x0 > x0 = [[fold#(/\x./\y.add(x, y), _x0, 0)]] 

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.