We consider the system Applicative_first_order_05__17.

  Alphabet:

    !facdot : [a * a] --> a 
    1 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    i : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    !facdot(1, x) => x 
    !facdot(x, 1) => x 
    !facdot(i(x), x) => 1 
    !facdot(x, i(x)) => 1 
    !facdot(i(x), !facdot(x, y)) => y 
    !facdot(x, !facdot(i(x), y)) => y 
    !facdot(!facdot(x, y), z) => !facdot(x, !facdot(y, z)) 
    i(1) => 1 
    i(i(x)) => x 
    i(!facdot(x, y)) => !facdot(i(y), i(x)) 
    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) 

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, all):

  Dependency Pairs P_0:

    0] !facdot#(!facdot(X, Y), Z) =#> !facdot#(X, !facdot(Y, Z))   
    1] !facdot#(!facdot(X, Y), Z) =#> !facdot#(Y, Z)   
    2] i#(!facdot(X, Y)) =#> !facdot#(i(Y), i(X))   
    3] i#(!facdot(X, Y)) =#> i#(Y)   
    4] i#(!facdot(X, Y)) =#> i#(X)   
    5] map#(F, cons(X, Y)) =#> map#(F, Y)   
    6] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    7] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    8] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    !facdot(1, X) => X 
    !facdot(X, 1) => X 
    !facdot(i(X), X) => 1 
    !facdot(X, i(X)) => 1 
    !facdot(i(X), !facdot(X, Y)) => Y 
    !facdot(X, !facdot(i(X), Y)) => Y 
    !facdot(!facdot(X, Y), Z) => !facdot(X, !facdot(Y, Z)) 
    i(1) => 1 
    i(i(X)) => X 
    i(!facdot(X, Y)) => !facdot(i(Y), i(X)) 
    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, all) is finite.

We consider the dependency pair problem (P_0, R_0, static, all).

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 : 0, 1 
    * 2 : 0, 1 
    * 3 : 2, 3, 4 
    * 4 : 2, 3, 4 
    * 5 : 5 
    * 6 : 7, 8 
    * 7 : 6 
    * 8 : 6 

This graph has the following strongly connected components:

  P_1:

    !facdot#(!facdot(X, Y), Z) =#> !facdot#(X, !facdot(Y, Z))   
    !facdot#(!facdot(X, Y), Z) =#> !facdot#(Y, Z)   

  P_2:

    i#(!facdot(X, Y)) =#> i#(Y)   
    i#(!facdot(X, Y)) =#> i#(X)   

  P_3:

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

  P_4:

    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), (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, all), (P_2, R_0, static, all), (P_3, R_0, static, all) and (P_4, R_0, static, all) is finite.

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

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_4, R_0, static, f) by (P_5, R_0, static, f), where P_5 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, all), (P_2, R_0, static, all), (P_3, R_0, static, all) and (P_5, R_0, static, all) is finite.

We consider the dependency pair problem (P_5, R_0, static, all).

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 each of (P_1, R_0, static, all), (P_2, R_0, static, all) and (P_3, R_0, static, all) is finite.

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

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_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, all) and (P_2, R_0, static, all) is finite.

We consider the dependency pair problem (P_2, R_0, static, all).

We apply the subterm criterion with the following projection function:

  nu(i#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(i#(!facdot(X, Y))) = !facdot(X, Y) |> Y = nu(i#(Y)) 
  nu(i#(!facdot(X, Y))) = !facdot(X, Y) |> X = nu(i#(X)) 

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, all) is finite.

We consider the dependency pair problem (P_1, R_0, static, all).

We apply the subterm criterion with the following projection function:

  nu(!facdot#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(!facdot#(!facdot(X, Y), Z)) = !facdot(X, Y) |> X = nu(!facdot#(X, !facdot(Y, Z))) 
  nu(!facdot#(!facdot(X, Y), Z)) = !facdot(X, Y) |> Y = nu(!facdot#(Y, Z)) 

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.