We consider the system Applicative_first_order_05__#3.38.

  Alphabet:

    0 : [] --> c 
    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 
    rev : [d] --> d 
    rev1 : [c * d] --> c 
    rev2 : [c * d] --> d 
    s : [a] --> c 
    true : [] --> b 

  Rules:

    rev(nil) => nil 
    rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) 
    rev1(0, nil) => 0 
    rev1(s(x), nil) => s(x) 
    rev1(x, cons(y, z)) => rev1(y, z) 
    rev2(x, nil) => nil 
    rev2(x, cons(y, z)) => rev(cons(x, rev2(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] rev#(cons(X, Y)) =#> rev1#(X, Y)   
    1] rev#(cons(X, Y)) =#> rev2#(X, Y)   
    2] rev1#(X, cons(Y, Z)) =#> rev1#(Y, Z)   
    3] rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(Y, Z)))   
    4] rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z)   
    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:

    rev(nil) => nil 
    rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
    rev1(0, nil) => 0 
    rev1(s(X), nil) => s(X) 
    rev1(X, cons(Y, Z)) => rev1(Y, Z) 
    rev2(X, nil) => nil 
    rev2(X, cons(Y, Z)) => rev(cons(X, rev2(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 : 2 
    * 1 : 3, 4 
    * 2 : 2 
    * 3 : 0, 1 
    * 4 : 3, 4 
    * 5 : 5 
    * 6 : 7, 8 
    * 7 : 6 
    * 8 : 6 

This graph has the following strongly connected components:

  P_1:

    rev#(cons(X, Y)) =#> rev2#(X, Y)   
    rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(Y, Z)))   
    rev2#(X, cons(Y, Z)) =#> rev2#(Y, Z)   

  P_2:

    rev1#(X, cons(Y, Z)) =#> rev1#(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(rev1#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(rev1#(X, cons(Y, Z))) = cons(Y, Z) |> Z = nu(rev1#(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 ::=

  rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
  rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  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) 

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 with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_1, R_1) are:

  rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) 
  rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, Z))) 

It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  rev#(cons(X, Y)) >? rev2#(X, Y) 
  rev2#(X, cons(Y, Z)) >? rev#(cons(X, rev2(Y, Z))) 
  rev2#(X, cons(Y, Z)) >? rev2#(Y, Z) 
  rev(cons(X, Y)) >= cons(rev1(X, Y), rev2(X, Y)) 
  rev2(X, cons(Y, Z)) >= rev(cons(X, rev2(Y, Z))) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y1 
  rev = \y0.y0 
  rev1 = \y0y1.0 
  rev2 = \y0y1.y1 
  rev2# = \y0y1.2y1 
  rev# = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[rev#(cons(_x0, _x1))]] = 4 + 2x1 > 2x1 = [[rev2#(_x0, _x1)]] 
  [[rev2#(_x0, cons(_x1, _x2))]] = 4 + 2x2 >= 4 + 2x2 = [[rev#(cons(_x0, rev2(_x1, _x2)))]] 
  [[rev2#(_x0, cons(_x1, _x2))]] = 4 + 2x2 > 2x2 = [[rev2#(_x1, _x2)]] 
  [[rev(cons(_x0, _x1))]] = 2 + x1 >= 2 + x1 = [[cons(rev1(_x0, _x1), rev2(_x0, _x1))]] 
  [[rev2(_x0, cons(_x1, _x2))]] = 2 + x2 >= 2 + x2 = [[rev(cons(_x0, rev2(_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:

  rev2#(X, cons(Y, Z)) =#> rev#(cons(X, rev2(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 place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:

    * 0 :  

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

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.