We consider the system Applicative_first_order_05__13.

  Alphabet:

    !facplus : [a * a] --> a 
    !factimes : [a * a] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) 
    !factimes(!facplus(x, y), z) => !facplus(!factimes(z, x), !factimes(z, y)) 
    !factimes(!factimes(x, y), z) => !factimes(x, !factimes(y, z)) 
    !facplus(!facplus(x, y), z) => !facplus(x, !facplus(y, z)) 
    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 [FuhKop19]).

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

  Dependency Pairs P_0:

    0] !factimes#(X, !facplus(Y, Z)) =#> !facplus#(!factimes(X, Y), !factimes(X, Z))   
    1] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y)   
    2] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z)   
    3] !factimes#(!facplus(X, Y), Z) =#> !facplus#(!factimes(Z, X), !factimes(Z, Y))   
    4] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X)   
    5] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y)   
    6] !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z))   
    7] !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z)   
    8] !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z))   
    9] !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z)   
    10] map#(F, cons(X, Y)) =#> map#(F, Y)   
    11] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    12] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    13] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) 
    !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) 
    !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) 
    !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 
    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, computable, formative) is finite.

We consider the dependency pair problem (P_0, R_0, computable, 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 : 8, 9 
    * 1 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 2 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 3 : 8, 9 
    * 4 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 5 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 6 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 7 : 0, 1, 2, 3, 4, 5, 6, 7 
    * 8 : 8, 9 
    * 9 : 8, 9 
    * 10 : 10 
    * 11 : 12, 13 
    * 12 : 11 
    * 13 : 11 

This graph has the following strongly connected components:

  P_1:

    !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y)   
    !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z)   
    !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X)   
    !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y)   
    !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z))   
    !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z)   

  P_2:

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

  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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_4, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_4, R_0, computable, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_4, R_0, computable, f) by (P_5, R_0, computable, 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, computable, formative), (P_2, R_0, computable, formative), (P_3, R_0, computable, formative) and (P_5, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_5, R_0, computable, 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 each of (P_1, R_0, computable, formative), (P_2, R_0, computable, formative) and (P_3, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_3, R_0, computable, 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 [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_0, computable, f) by ({}, R_0, computable, 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, computable, formative) and (P_2, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_2, R_0, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(!facplus#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_2, R_0, computable, f) by ({}, R_0, computable, 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, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_0, computable, formative).

The formative rules of (P_1, R_0) are R_1 ::=

  !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) 
  !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) 
  !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) 
  !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, computable, formative) by (P_1, R_1, computable, formative).

Thus, the original system is terminating if (P_1, R_1, computable, formative) is finite.

We consider the dependency pair problem (P_1, R_1, computable, 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:

  !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Y) 
  !factimes#(X, !facplus(Y, Z)) >? !factimes#(X, Z) 
  !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, X) 
  !factimes#(!facplus(X, Y), Z) >? !factimes#(Z, Y) 
  !factimes#(!factimes(X, Y), Z) >? !factimes#(X, !factimes(Y, Z)) 
  !factimes#(!factimes(X, Y), Z) >? !factimes#(Y, Z) 
  !factimes(X, !facplus(Y, Z)) >= !facplus(!factimes(X, Y), !factimes(X, Z)) 
  !factimes(!facplus(X, Y), Z) >= !facplus(!factimes(Z, X), !factimes(Z, Y)) 
  !factimes(!factimes(X, Y), Z) >= !factimes(X, !factimes(Y, Z)) 
  !facplus(!facplus(X, Y), Z) >= !facplus(X, !facplus(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !facplus = \y0y1.2 + y1 + 2y0 
  !factimes = \y0y1.y0 + y1 + 2y0y1 
  !factimes# = \y0y1.y0 + y1 + 2y0y1 

Using this interpretation, the requirements translate to:

  [[!factimes#(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x2 + 2x1 + 4x0x1 + 5x0 > x0 + x1 + 2x0x1 = [[!factimes#(_x0, _x1)]] 
  [[!factimes#(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x2 + 2x1 + 4x0x1 + 5x0 > x0 + x2 + 2x0x2 = [[!factimes#(_x0, _x2)]] 
  [[!factimes#(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x1x2 + 4x0x2 + 5x2 > x0 + x2 + 2x0x2 = [[!factimes#(_x2, _x0)]] 
  [[!factimes#(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x1x2 + 4x0x2 + 5x2 > x1 + x2 + 2x1x2 = [[!factimes#(_x2, _x1)]] 
  [[!factimes#(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 >= x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 = [[!factimes#(_x0, !factimes(_x1, _x2))]] 
  [[!factimes#(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 >= x1 + x2 + 2x1x2 = [[!factimes#(_x1, _x2)]] 
  [[!factimes(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x2 + 2x1 + 4x0x1 + 5x0 >= 2 + x2 + 2x0x2 + 2x1 + 3x0 + 4x0x1 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[!factimes(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x1x2 + 4x0x2 + 5x2 >= 2 + x1 + 2x0 + 2x1x2 + 3x2 + 4x0x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] 
  [[!factimes(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 >= x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] 
  [[!facplus(!facplus(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 >= 4 + x2 + 2x0 + 2x1 = [[!facplus(_x0, !facplus(_x1, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_1, computable, formative) by (P_6, R_1, computable, formative), where P_6 consists of:

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

Thus, the original system is terminating if (P_6, R_1, computable, formative) is finite.

We consider the dependency pair problem (P_6, R_1, computable, formative).

We apply the subterm criterion with the following projection function:

  nu(!factimes#) = 1 

Thus, we can orient the dependency pairs as follows:

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

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_6, R_1, computable, f) by ({}, R_1, computable, 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 +++

[FuhKop19]  C. Fuhs, and C. Kop.  A static higher-order dependency pair framework.  In Proceedings of ESOP 2019, 2019.
[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[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.