We consider the system Applicative_first_order_05__18.

  Alphabet:

    !facplus : [b * b] --> b 
    !factimes : [a * b] --> b 
    cons : [d * e] --> e 
    false : [] --> c 
    filter : [d -> c * e] --> e 
    filter2 : [c * d -> c * d * e] --> e 
    map : [d -> d * e] --> e 
    nil : [] --> e 
    true : [] --> c 

  Rules:

    !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, 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 rule removal, following [Kop12, Theorem 2.23].

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

  !factimes(X, !facplus(Y, Z)) >? !facplus(!factimes(X, Y), !factimes(X, 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) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 1 + x1 + x2 + 2x0 + x0x1 + x0x2 >= 1 + x1 + x2 + 2x0 + x0x1 + x0x2 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[map(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + F0(0) + 3x1F0(1 + x1 + x2) + 3x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > 1 + x1 + 2x2 + F0(0) + F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 3x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 3 + x1 + 3x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 2 + 3x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  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 rule removal, following [Kop12, Theorem 2.23].

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

  !factimes(X, !facplus(Y, Z)) >? !facplus(!factimes(X, Y), !factimes(X, Z)) 
  map(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  !facplus = \y0y1.1 + y0 + y1 
  !factimes = \y0y1.y0 + y1 + y0y1 
  map = \G0y1.3 + 3y1 + G0(0) 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 1 + x1 + x2 + 2x0 + x0x1 + x0x2 >= 1 + x1 + x2 + 2x0 + x0x1 + x0x2 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  map(F, nil) => nil 

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

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

  !factimes(X, !facplus(Y, Z)) >? !facplus(!factimes(X, Y), !factimes(X, Z)) 

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

The following interpretation satisfies the requirements:

  !facplus = \y0y1.2 + y0 + y1 
  !factimes = \y0y1.y0 + 3y1 + y0y1 

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 6 + 3x0 + 3x1 + 3x2 + x0x1 + x0x2 > 2 + 2x0 + 3x1 + 3x2 + x0x1 + x0x2 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 

We can thus remove the following rules:

  !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) 

All rules were succesfully removed.  Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.