We consider the system AotoYamada_05__027.

  Alphabet:

    0 : [] --> a 
    cons : [a * b] --> b 
    inc : [b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 
    plus : [a] --> a -> a 
    s : [a] --> a 

  Rules:

    plus(0) x => x 
    plus(s(x)) y => s(plus(x) y) 
    inc(x) => map(plus(s(0)), x) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(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]).

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)) 
  inc(X) => map(/\x.plus(s(0), x), X) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, 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] inc#(X) =#> map#(/\x.plus(s(0), x), X)   
    2] inc#(X) =#> plus#(s(0), Y)   
    3] map#(F, cons(X, Y)) =#> map#(F, Y)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    inc(X) => map(/\x.plus(s(0), x), X) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(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 : 3 
    * 2 : 0 
    * 3 : 3 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    map#(F, cons(X, Y)) =#> map#(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 [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  map#(F, cons(X, Y)) >? map#(F, Y) 
  plus(0, X) >= X 
  plus(s(X), Y) >= s(plus(X, Y)) 
  inc(X) >= map(/\x.plus(s(0), x), X) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

  pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-inc# = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.2 + y1 
  inc = \y0.0 
  map = \G0y1.y1 
  map# = \G0y1.y1 
  nil = 0 
  plus = \y0y1.3 + 2y1 
  s = \y0.0 

Using this interpretation, the requirements translate to:

  [[map#(_F0, cons(_x1, _x2))]] = 2 + x2 > x2 = [[map#(_F0, _x2)]] 
  [[plus(0, _x0)]] = 3 + 2x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 3 + 2x1 >= 0 = [[s(plus(_x0, _x1))]] 
  [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 3 + x0 >= x0 = [[map(/\x.plus(s(0), x), _x0)]] 
  [[map(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[cons(_F0 _x1, map(_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 [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)) 
  inc(X) >= map(/\x.plus(s(0), x), X) 
  map(F, nil) >= nil 
  map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 

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

  pi( inc(X) ) = #argfun-inc#(map(/\x.plus(s(0), x), X)) 

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

The following interpretation satisfies the requirements:

  #argfun-inc# = \y0.3 + y0 
  0 = 0 
  cons = \y0y1.3 
  inc = \y0.0 
  map = \G0y1.3 + 2G0(0) + 3y1G0(y1) 
  nil = 0 
  plus = \y0y1.y1 + 3y0 
  plus# = \y0y1.y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[plus#(s(_x0), _x1)]] = 3 + x0 > x0 = [[plus#(_x0, _x1)]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 9 + x1 + 3x0 >= 3 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
  [[#argfun-inc#(map(/\x.plus(s(0), x), _x0))]] = 24 + 3x0x0 + 27x0 >= 21 + 3x0x0 + 27x0 = [[map(/\x.plus(s(0), x), _x0)]] 
  [[map(_F0, nil)]] = 3 + 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 3 + 2F0(0) + 9F0(3) >= 3 = [[cons(_F0 _x1, map(_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.