We consider the system AotoYamada_05__013.

  Alphabet:

    append : [c * c] --> c 
    cons : [b * c] --> c 
    flatwith : [a -> b * b] --> c 
    flatwithsub : [a -> b * c] --> c 
    leaf : [a] --> b 
    nil : [] --> c 
    node : [c] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    flatwith(f, leaf(x)) => cons(f x, nil) 
    flatwith(f, node(x)) => flatwithsub(f, x) 
    flatwithsub(f, nil) => nil 
    flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(f, y)) 

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] append#(cons(X, Y), Z) =#> append#(Y, Z)   
    1] flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    2] flatwithsub#(F, cons(X, Y)) =#> append#(flatwith(F, X), flatwithsub(F, Y))   
    3] flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    4] flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    flatwith(F, leaf(X)) => cons(F X, nil) 
    flatwith(F, node(X)) => flatwithsub(F, X) 
    flatwithsub(F, nil) => nil 
    flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(F, 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, 3, 4 
    * 2 : 0 
    * 3 : 1 
    * 4 : 2, 3, 4 

This graph has the following strongly connected components:

  P_1:

    append#(cons(X, Y), Z) =#> append#(Y, Z)   

  P_2:

    flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, 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) and (P_2, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  (P_2, R_0) has no usable rules.

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

  flatwith#(F, node(X)) >? flatwithsub#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwith#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.3 + y0 + 2y1 
  flatwith# = \G0y1.1 + y1 
  flatwithsub# = \G0y1.y1 
  node = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[flatwith#(_F0, node(_x1))]] = 4 + x1 > x1 = [[flatwithsub#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 + x1 + 2x2 > 1 + x1 = [[flatwith#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 + x1 + 2x2 > x2 = [[flatwithsub#(_F0, _x2)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

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:

  append#(cons(X, Y), Z) >? append#(Y, Z) 

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

The following interpretation satisfies the requirements:

  append# = \y0y1.y0 
  cons = \y0y1.1 + 2y1 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 1 + 2x1 > x1 = [[append#(_x1, _x2)]] 

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.