We consider the system Applicative_05__Ex7_9.

  Alphabet:

    0 : [] --> a 
    cons : [b * c] --> c 
    d : [a * a] --> c 
    false : [] --> c 
    filter : [b -> c * c] --> c 
    gtr : [a * a] --> c 
    if : [c * c * c] --> c 
    len : [c] --> a 
    nil : [] --> c 
    s : [a] --> a 
    sub : [a * a] --> a 
    true : [] --> c 

  Rules:

    if(true, x, y) => x 
    if(false, x, y) => y 
    sub(x, 0) => x 
    sub(s(x), s(y)) => sub(x, y) 
    gtr(0, x) => false 
    gtr(s(x), 0) => true 
    gtr(s(x), s(y)) => gtr(x, y) 
    d(x, 0) => true 
    d(s(x), s(y)) => if(gtr(x, y), false, d(s(x), sub(y, x))) 
    len(nil) => 0 
    len(cons(x, y)) => s(len(y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => if(f x, cons(x, filter(f, 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] sub#(s(X), s(Y)) =#> sub#(X, Y)   
    1] gtr#(s(X), s(Y)) =#> gtr#(X, Y)   
    2] d#(s(X), s(Y)) =#> if#(gtr(X, Y), false, d(s(X), sub(Y, X)))   
    3] d#(s(X), s(Y)) =#> gtr#(X, Y)   
    4] d#(s(X), s(Y)) =#> d#(s(X), sub(Y, X))   
    5] d#(s(X), s(Y)) =#> sub#(Y, X)   
    6] len#(cons(X, Y)) =#> len#(Y)   
    7] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y))   
    8] filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    9] filter#(F, cons(X, Y)) =#> filter#(F, Y)   

  Rules R_0:

    if(true, X, Y) => X 
    if(false, X, Y) => Y 
    sub(X, 0) => X 
    sub(s(X), s(Y)) => sub(X, Y) 
    gtr(0, X) => false 
    gtr(s(X), 0) => true 
    gtr(s(X), s(Y)) => gtr(X, Y) 
    d(X, 0) => true 
    d(s(X), s(Y)) => if(gtr(X, Y), false, d(s(X), sub(Y, X))) 
    len(nil) => 0 
    len(cons(X, Y)) => s(len(Y)) 
    filter(F, nil) => nil 
    filter(F, cons(X, Y)) => if(F X, cons(X, filter(F, 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 : 1 
    * 2 :  
    * 3 : 1 
    * 4 : 2, 3, 4, 5 
    * 5 : 0 
    * 6 : 6 
    * 7 :  
    * 8 : 7, 8, 9 
    * 9 : 7, 8, 9 

This graph has the following strongly connected components:

  P_1:

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

  P_2:

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

  P_3:

    d#(s(X), s(Y)) =#> d#(s(X), sub(Y, X))   

  P_4:

    len#(cons(X, Y)) =#> len#(Y)   

  P_5:

    filter#(F, cons(X, Y)) =#> filter#(F, Y)   
    filter#(F, cons(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), (P_4, R_0, m, f) and (P_5, 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), (P_4, 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 apply the subterm criterion with the following projection function:

  nu(filter#) = 2 

Thus, we can orient the dependency pairs as follows:

  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 
  nu(filter#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(filter#(F, Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_5, 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), (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(len#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(len#(cons(X, Y))) = cons(X, Y) |> Y = nu(len#(Y)) 

By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_4, 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), (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 will use the reduction pair processor with usable rules [Kop12, Thm. 7.44].  The usable rules of (P_3, R_0) are:

  sub(X, 0) => X 
  sub(s(X), s(Y)) => sub(X, Y) 

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

  d#(s(X), s(Y)) >? d#(s(X), sub(Y, X)) 
  sub(X, 0) >= X 
  sub(s(X), s(Y)) >= sub(X, Y) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  d# = \y0y1.3y1 
  s = \y0.3 + 3y0 
  sub = \y0y1.y0 

Using this interpretation, the requirements translate to:

  [[d#(s(_x0), s(_x1))]] = 9 + 9x1 > 3x1 = [[d#(s(_x0), sub(_x1, _x0))]] 
  [[sub(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[sub(s(_x0), s(_x1))]] = 3 + 3x0 >= x0 = [[sub(_x0, _x1)]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  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(gtr#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(gtr#(s(X), s(Y))) = s(X) |> X = nu(gtr#(X, 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(sub#) = 1 

Thus, we can orient the dependency pairs as follows:

  nu(sub#(s(X), s(Y))) = s(X) |> X = nu(sub#(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.