We consider the system AotoYamada_05__022.

  Alphabet:

    cons : [b * c] --> c 
    leaf : [a] --> b 
    mapt : [a -> a * b] --> b 
    maptlist : [a -> a * c] --> c 
    nil : [] --> c 
    node : [c] --> b 

  Rules:

    mapt(f, leaf(x)) => leaf(f x) 
    mapt(f, node(x)) => node(maptlist(f, x)) 
    maptlist(f, nil) => nil 
    maptlist(f, cons(x, y)) => cons(mapt(f, x), maptlist(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] mapt#(F, node(X)) =#> maptlist#(F, X)   
    1] maptlist#(F, cons(X, Y)) =#> mapt#(F, X)   
    2] maptlist#(F, cons(X, Y)) =#> maptlist#(F, Y)   

  Rules R_0:

    mapt(F, leaf(X)) => leaf(F X) 
    mapt(F, node(X)) => node(maptlist(F, X)) 
    maptlist(F, nil) => nil 
    maptlist(F, cons(X, Y)) => cons(mapt(F, X), maptlist(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 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:

  mapt#(F, node(X)) >? maptlist#(F, X) 
  maptlist#(F, cons(X, Y)) >? mapt#(F, X) 
  maptlist#(F, cons(X, Y)) >? maptlist#(F, Y) 
  mapt(F, leaf(X)) >= leaf(F X) 
  mapt(F, node(X)) >= node(maptlist(F, X)) 
  maptlist(F, nil) >= nil 
  maptlist(F, cons(X, Y)) >= cons(mapt(F, X), maptlist(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y0 + 2y1 
  leaf = \y0.0 
  mapt = \G0y1.2 + 2y1 + 2y1G0(y1) 
  mapt# = \G0y1.2y1 
  maptlist = \G0y1.2y1 + 2y1G0(y1) 
  maptlist# = \G0y1.2y1 
  nil = 0 
  node = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[mapt#(_F0, node(_x1))]] = 4x1 >= 2x1 = [[maptlist#(_F0, _x1)]] 
  [[maptlist#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 4x2 > 2x1 = [[mapt#(_F0, _x1)]] 
  [[maptlist#(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 4x2 > 2x2 = [[maptlist#(_F0, _x2)]] 
  [[mapt(_F0, leaf(_x1))]] = 2 >= 0 = [[leaf(_F0 _x1)]] 
  [[mapt(_F0, node(_x1))]] = 2 + 4x1 + 4x1F0(2x1) >= 4x1 + 4x1F0(x1) = [[node(maptlist(_F0, _x1))]] 
  [[maptlist(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[maptlist(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 4x2 + 2x1F0(2 + x1 + 2x2) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) >= 4 + 2x1 + 4x2 + 2x1F0(x1) + 4x2F0(x2) = [[cons(mapt(_F0, _x1), maptlist(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

  mapt#(F, node(X)) =#> maptlist#(F, X)   

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:

  mapt#(F, node(X)) >? maptlist#(F, X) 
  mapt(F, leaf(X)) >= leaf(F X) 
  mapt(F, node(X)) >= node(maptlist(F, X)) 
  maptlist(F, nil) >= nil 
  maptlist(F, cons(X, Y)) >= cons(mapt(F, X), maptlist(F, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.0 
  leaf = \y0.0 
  mapt = \G0y1.2 + 2G0(0) + 3G0(y1) 
  mapt# = \G0y1.3 
  maptlist = \G0y1.2 
  maptlist# = \G0y1.0 
  nil = 0 
  node = \y0.0 

Using this interpretation, the requirements translate to:

  [[mapt#(_F0, node(_x1))]] = 3 > 0 = [[maptlist#(_F0, _x1)]] 
  [[mapt(_F0, leaf(_x1))]] = 2 + 5F0(0) >= 0 = [[leaf(_F0 _x1)]] 
  [[mapt(_F0, node(_x1))]] = 2 + 5F0(0) >= 0 = [[node(maptlist(_F0, _x1))]] 
  [[maptlist(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[maptlist(_F0, cons(_x1, _x2))]] = 2 >= 0 = [[cons(mapt(_F0, _x1), maptlist(_F0, _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.