We consider the system Applicative_first_order_05__#3.32.

  Alphabet:

    a : [] --> a 
    b : [] --> a 
    cons : [e * f] --> f 
    f : [a] --> b 
    false : [] --> d 
    filter : [e -> d * f] --> f 
    filter2 : [d * e -> d * e * f] --> f 
    g : [a] --> c 
    map : [e -> e * f] --> f 
    nil : [] --> f 
    true : [] --> d 

  Rules:

    f(a) => f(b) 
    g(b) => g(a) 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, 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] f#(a) =#> f#(b)   
    1] g#(b) =#> g#(a)   
    2] map#(F, cons(X, Y)) =#> map#(F, Y)   
    3] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    4] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    5] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    f(a) => f(b) 
    g(b) => g(a) 
    map(F, nil) => nil 
    map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
    filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
    filter2(false, F, X, Y) => filter(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 :  
    * 1 :  
    * 2 : 2 
    * 3 : 4, 5 
    * 4 : 3 
    * 5 : 3 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

    filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    filter2#(true, F, X, Y) =#> filter#(F, Y)   
    filter2#(false, F, X, Y) =#> filter#(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(filter2#) = 4 
  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter2#(F X, F, X, Y)) 
  nu(filter2#(true, F, X, Y)) = Y = Y = nu(filter#(F, Y)) 
  nu(filter2#(false, F, X, Y)) = Y = Y = nu(filter#(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 (P_3, R_0, static, f), where P_3 contains:

  filter2#(true, F, X, Y) =#> filter#(F, Y)   
  filter2#(false, F, X, Y) =#> filter#(F, Y)   

Thus, the original system is terminating if each of (P_1, 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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  
    * 1 :  

This graph has no strongly connected components.  By [Kop12, Thm. 7.31], this implies finiteness of the dependency pair problem.

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(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_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.