We consider the system Applicative_05__Ex5Sorting.

  Alphabet:

    0 : [] --> a 
    ascending!fac6220sort : [b] --> b 
    cons : [a * b] --> b 
    descending!fac6220sort : [b] --> b 
    insert : [a -> a -> a * a -> a -> a * b * a] --> b 
    max : [] --> a -> a -> a 
    min : [] --> a -> a -> a 
    nil : [] --> b 
    s : [a] --> a 
    sort : [a -> a -> a * a -> a -> a * b] --> b 

  Rules:

    max 0 x => x 
    max x 0 => x 
    max s(x) s(y) => max x y 
    min 0 x => 0 
    min x 0 => 0 
    min s(x) s(y) => min x y 
    insert(f, g, nil, x) => cons(x, nil) 
    insert(f, g, cons(x, y), z) => cons(f z x, insert(f, g, y, g z x)) 
    sort(f, g, nil) => nil 
    sort(f, g, cons(x, y)) => insert(f, g, sort(f, g, y), x) 
    ascending!fac6220sort(x) => sort(min, max, x) 
    descending!fac6220sort(x) => sort(max, min, 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]).

In order to do so, we start by eta-expanding the system, which gives:

  max(0, X) => X 
  max(X, 0) => X 
  max(s(X), s(Y)) => max(X, Y) 
  min(0, X) => 0 
  min(X, 0) => 0 
  min(s(X), s(Y)) => min(X, Y) 
  insert(F, G, nil, X) => cons(X, nil) 
  insert(F, G, cons(X, Y), Z) => cons(F Z X, insert(F, G, Y, G Z X)) 
  sort(F, G, nil) => nil 
  sort(F, G, cons(X, Y)) => insert(F, G, sort(F, G, Y), X) 
  ascending!fac6220sort(X) => sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X) 
  descending!fac6220sort(X) => sort(/\x./\y.max(x, y), /\z./\u.min(z, u), X) 

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] max#(s(X), s(Y)) =#> max#(X, Y)   
    1] min#(s(X), s(Y)) =#> min#(X, Y)   
    2] insert#(F, G, cons(X, Y), Z) =#> insert#(F, G, Y, G Z X)   
    3] sort#(F, G, cons(X, Y)) =#> insert#(F, G, sort(F, G, Y), X)   
    4] sort#(F, G, cons(X, Y)) =#> sort#(F, G, Y)   
    5] ascending!fac6220sort#(X) =#> sort#(/\x./\y.min(x, y), /\z./\u.max(z, u), X)   
    6] ascending!fac6220sort#(X) =#> min#(Y, Z)   
    7] ascending!fac6220sort#(X) =#> max#(Y, Z)   
    8] descending!fac6220sort#(X) =#> sort#(/\x./\y.max(x, y), /\z./\u.min(z, u), X)   
    9] descending!fac6220sort#(X) =#> max#(Y, Z)   
    10] descending!fac6220sort#(X) =#> min#(Y, Z)   

  Rules R_0:

    max(0, X) => X 
    max(X, 0) => X 
    max(s(X), s(Y)) => max(X, Y) 
    min(0, X) => 0 
    min(X, 0) => 0 
    min(s(X), s(Y)) => min(X, Y) 
    insert(F, G, nil, X) => cons(X, nil) 
    insert(F, G, cons(X, Y), Z) => cons(F Z X, insert(F, G, Y, G Z X)) 
    sort(F, G, nil) => nil 
    sort(F, G, cons(X, Y)) => insert(F, G, sort(F, G, Y), X) 
    ascending!fac6220sort(X) => sort(/\x./\y.min(x, y), /\z./\u.max(z, u), X) 
    descending!fac6220sort(X) => sort(/\x./\y.max(x, y), /\z./\u.min(z, u), 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 : 2 
    * 3 : 2 
    * 4 : 3, 4 
    * 5 : 3, 4 
    * 6 : 1 
    * 7 : 0 
    * 8 : 3, 4 
    * 9 : 0 
    * 10 : 1 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

    insert#(F, G, cons(X, Y), Z) =#> insert#(F, G, Y, G Z X)   

  P_4:

    sort#(F, G, cons(X, Y)) =#> sort#(F, G, 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), (P_2, R_0, m, f), (P_3, R_0, m, f) and (P_4, R_0, m, f).

Thus, the original system is terminating if each of (P_1, R_0, static, formative), (P_2, R_0, static, formative), (P_3, R_0, static, formative) and (P_4, R_0, static, formative) is finite.

We consider the dependency pair problem (P_4, R_0, static, formative).

We apply the subterm criterion with the following projection function:

  nu(sort#) = 3 

Thus, we can orient the dependency pairs as follows:

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

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_4, 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 each of (P_1, R_0, static, formative), (P_2, R_0, static, formative) and (P_3, R_0, static, formative) is finite.

We consider the dependency pair problem (P_3, R_0, static, formative).

We apply the subterm criterion with the following projection function:

  nu(insert#) = 3 

Thus, we can orient the dependency pairs as follows:

  nu(insert#(F, G, cons(X, Y), Z)) = cons(X, Y) |> Y = nu(insert#(F, G, Y, G Z X)) 

By [Kop12, Thm. 7.35] and [Kop13, Thm. 5], we may replace a dependency pair problem (P_3, 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 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(min#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(min#(s(X), s(Y))) = s(X) |> X = nu(min#(X, 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(max#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(max#(s(X), s(Y))) = s(X) |> X = nu(max#(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.