We consider the system AotoYamada_05__011.

  Alphabet:

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

  Rules:

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

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)) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  curry(F, X, Y) => F X Y 
  inc(X) => map(/\x.curry(/\y./\z.plus(y, z), s(0), x), X) 

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] map#(F, cons(X, Y)) =#> map#(F, Y)   
    2] inc#(X) =#> map#(/\x.curry(/\y./\z.plus(y, z), s(0), x), X)   
    3] inc#(X) =#> curry#(/\x./\y.plus(x, y), s(0), Y)   
    4] inc#(X) =#> plus#(Y, Z)   

  Rules R_0:

    plus(0, X) => X 
    plus(s(X), Y) => s(plus(X, Y)) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    curry(F, X, Y) => F X Y 
    inc(X) => map(/\x.curry(/\y./\z.plus(y, z), s(0), x), 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 : 1 
    * 2 : 1 
    * 3 :  
    * 4 : 0 

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 apply the subterm criterion with the following projection function:

  nu(map#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_2, R_0, static, f) by ({}, R_0, static, f).  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 apply the subterm criterion with the following projection function:

  nu(plus#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_1, R_0, static, f) by ({}, R_0, static, f).  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.