We consider the system Applicative_first_order_05__#3.25.

  Alphabet:

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

  Rules:

    f(g(x)) => g(f(f(x))) 
    f(h(x)) => h(g(x)) 
    map(i, nil) => nil 
    map(i, cons(x, y)) => cons(i x, map(i, y)) 
    filter(i, nil) => nil 
    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
    filter2(true, i, x, y) => cons(x, filter(i, y)) 
    filter2(false, i, x, y) => filter(i, y) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  f(g(X)) >? g(f(f(X))) 
  f(h(X)) >? h(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) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.1 + y0 + y1 
  f = \y0.y0 
  false = 3 
  filter = \G0y1.2 + 2y1 + G0(0) + 2y1G0(y1) 
  filter2 = \y0G1y2y3.1 + y0 + y2 + 2y3 + G1(0) + 2y3G1(y3) 
  g = \y0.y0 
  h = \y0.y0 
  map = \G0y1.2 + 3y1 + 2y1G0(y1) + 2G0(y1) 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] 
  [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] 
  [[map(_F0, nil)]] = 2 + 2F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 4F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 1 + 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  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) 

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#(g(X)) =#> f#(f(X))   
    1] f#(g(X)) =#> f#(X)   

  Rules R_0:

    f(g(X)) => g(f(f(X))) 
    f(h(X)) => h(g(X)) 

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).

The formative rules of (P_0, R_0) are R_1 ::=

  f(g(X)) => g(f(f(X))) 

By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, computable, formative) by (P_0, R_1, computable, formative).

Thus, the original system is terminating if (P_0, R_1, computable, formative) is finite.

We consider the dependency pair problem (P_0, 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#(g(X)) >? f#(f(X)) 
  f#(g(X)) >? f#(X) 
  f(g(X)) >= g(f(f(X))) 

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

We consider usable rules with respect to the following argument filtering:


This leaves the following ordering requirements:

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

The following interpretation satisfies the requirements:

  f = \y0.y0 
  f# = \y0.3y0 
  g = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(f(_x0))]] 
  [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(_x0)]] 
  [[f(g(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[g(f(f(_x0)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1).  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.