We consider the system Applicative_first_order_05__hydra.

  Alphabet:

    0 : [] --> a 
    cons : [c * c] --> c 
    copy : [a * c * c] --> c 
    f : [c] --> c 
    false : [] --> b 
    filter : [c -> b * c] --> c 
    filter2 : [b * c -> b * c * c] --> c 
    map : [c -> c * c] --> c 
    n : [] --> a 
    nil : [] --> c 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    f(cons(nil, x)) => x 
    f(cons(f(cons(nil, x)), y)) => copy(n, x, y) 
    copy(0, x, y) => f(y) 
    copy(s(x), y, z) => copy(x, y, cons(f(y), z)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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] f#(cons(f(cons(nil, X)), Y)) =#> copy#(n, X, Y)   
    1] copy#(0, X, Y) =#> f#(Y)   
    2] copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z))   
    3] copy#(s(X), Y, Z) =#> f#(Y)   
    4] map#(F, cons(X, Y)) =#> map#(F, Y)   
    5] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    6] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    7] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    f(cons(nil, X)) => X 
    f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) 
    copy(0, X, Y) => f(Y) 
    copy(s(X), Y, Z) => copy(X, Y, cons(f(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 :  
    * 1 : 0 
    * 2 : 1, 2, 3 
    * 3 : 0 
    * 4 : 4 
    * 5 : 6, 7 
    * 6 : 5 
    * 7 : 5 

This graph has the following strongly connected components:

  P_1:

    copy#(s(X), Y, Z) =#> copy#(X, Y, cons(f(Y), Z))   

  P_2:

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

  P_3:

    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) and (P_3, 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) 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(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_3, R_0, computable, f) by (P_4, R_0, computable, f), where P_4 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) and (P_4, R_0, computable, formative) is finite.

We consider the dependency pair problem (P_4, 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) 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(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_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).

We apply the subterm criterion with the following projection function:

  nu(copy#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(copy#(s(X), Y, Z)) = s(X) |> X = nu(copy#(X, Y, cons(f(Y), Z))) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_1, R_0, computable, f) by ({}, R_0, 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.