We consider the system length.

  Alphabet:

    0 : [] --> nat 
    cons : [nat * list] --> list 
    foldr : [nat -> nat -> nat * nat * list] --> nat 
    length : [list] --> nat 
    nil : [] --> list 
    s : [nat] --> nat 

  Rules:

    foldr(f, x, nil) => x 
    foldr(f, x, cons(y, z)) => f y foldr(f, x, z) 
    length(x) => foldr(/\y./\z.s(z), 0, 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] foldr#(F, X, cons(Y, Z)) =#> foldr#(F, X, Z)   
    1] length#(X) =#> foldr#(/\x./\y.s(y), 0, X)   

  Rules R_0:

    foldr(F, X, nil) => X 
    foldr(F, X, cons(Y, Z)) => F Y foldr(F, X, Z) 
    length(X) => foldr(/\x./\y.s(y), 0, 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).

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 : 0 

This graph has the following strongly connected components:

  P_1:

    foldr#(F, X, cons(Y, Z)) =#> foldr#(F, X, Z)   

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 with usable rules [Kop12, Thm. 7.44].  (P_1, R_0) has no usable rules.

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  foldr#(F, X, cons(Y, Z)) >? foldr#(F, X, Z) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + 2y1 
  foldr# = \G0y1y2.y2 

Using this interpretation, the requirements translate to:

  [[foldr#(_F0, _x1, cons(_x2, _x3))]] = 3 + 2x3 > x3 = [[foldr#(_F0, _x1, _x3)]] 

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.