We consider the system Applicative_first_order_05__#3.36.

  Alphabet:

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

  Rules:

    minus(x, 0) => x 
    minus(s(x), s(y)) => minus(x, y) 
    f(0) => s(0) 
    f(s(x)) => minus(s(x), g(f(x))) 
    g(0) => 0 
    g(s(x)) => minus(s(x), f(g(x))) 
    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 [FuhKop19]).

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

  Dependency Pairs P_0:

    0] minus#(s(X), s(Y)) =#> minus#(X, Y)   
    1] f#(s(X)) =#> minus#(s(X), g(f(X)))   
    2] f#(s(X)) =#> g#(f(X))   
    3] f#(s(X)) =#> f#(X)   
    4] g#(s(X)) =#> minus#(s(X), f(g(X)))   
    5] g#(s(X)) =#> f#(g(X))   
    6] g#(s(X)) =#> g#(X)   
    7] map#(F, cons(X, Y)) =#> map#(F, Y)   
    8] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y)   
    9] filter2#(true, F, X, Y) =#> filter#(F, Y)   
    10] filter2#(false, F, X, Y) =#> filter#(F, Y)   

  Rules R_0:

    minus(X, 0) => X 
    minus(s(X), s(Y)) => minus(X, Y) 
    f(0) => s(0) 
    f(s(X)) => minus(s(X), g(f(X))) 
    g(0) => 0 
    g(s(X)) => minus(s(X), f(g(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, 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 : 0 
    * 1 : 0 
    * 2 : 4, 5, 6 
    * 3 : 1, 2, 3 
    * 4 : 0 
    * 5 : 1, 2, 3 
    * 6 : 4, 5, 6 
    * 7 : 7 
    * 8 : 9, 10 
    * 9 : 8 
    * 10 : 8 

This graph has the following strongly connected components:

  P_1:

    minus#(s(X), s(Y)) =#> minus#(X, Y)   

  P_2:

    f#(s(X)) =#> g#(f(X))   
    f#(s(X)) =#> f#(X)   
    g#(s(X)) =#> f#(g(X))   
    g#(s(X)) =#> g#(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, 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).

The formative rules of (P_2, R_0) are R_1 ::=

  minus(X, 0) => X 
  minus(s(X), s(Y)) => minus(X, Y) 
  f(0) => s(0) 
  f(s(X)) => minus(s(X), g(f(X))) 
  g(0) => 0 
  g(s(X)) => minus(s(X), f(g(X))) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, computable, formative) by (P_2, R_1, computable, formative).

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (P_2, R_1, computable, formative) is finite.

We consider the dependency pair problem (P_2, 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:

  f#(s(X)) >? g#(f(X)) 
  f#(s(X)) >? f#(X) 
  g#(s(X)) >? f#(g(X)) 
  g#(s(X)) >? g#(X) 
  minus(X, 0) >= X 
  minus(s(X), s(Y)) >= minus(X, Y) 
  f(0) >= s(0) 
  f(s(X)) >= minus(s(X), g(f(X))) 
  g(0) >= 0 
  g(s(X)) >= minus(s(X), f(g(X))) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0.1 + y0 
  f# = \y0.1 + y0 
  g = \y0.2y0 
  g# = \y0.2y0 
  minus = \y0y1.y0 
  s = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[f#(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[g#(f(_x0))]] 
  [[f#(s(_x0))]] = 2 + 2x0 > 1 + x0 = [[f#(_x0)]] 
  [[g#(s(_x0))]] = 2 + 4x0 > 1 + 2x0 = [[f#(g(_x0))]] 
  [[g#(s(_x0))]] = 2 + 4x0 > 2x0 = [[g#(_x0)]] 
  [[minus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[minus(s(_x0), s(_x1))]] = 1 + 2x0 >= x0 = [[minus(_x0, _x1)]] 
  [[f(0)]] = 1 >= 1 = [[s(0)]] 
  [[f(s(_x0))]] = 2 + 2x0 >= 1 + 2x0 = [[minus(s(_x0), g(f(_x0)))]] 
  [[g(0)]] = 0 >= 0 = [[0]] 
  [[g(s(_x0))]] = 2 + 4x0 >= 1 + 2x0 = [[minus(s(_x0), f(g(_x0)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_1, computable, formative) by (P_6, R_1, computable, formative), where P_6 consists of:

  f#(s(X)) =#> g#(f(X))   

Thus, the original system is terminating if each of (P_1, R_0, computable, formative) and (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.

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(minus#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(minus#(s(X), s(Y))) = s(X) |> X = nu(minus#(X, Y)) 

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.