We consider the system Applicative_first_order_05__13.

  Alphabet:

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

  Rules:

    !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) 
    !factimes(!facplus(x, y), z) => !facplus(!factimes(z, x), !factimes(z, y)) 
    !factimes(!factimes(x, y), z) => !factimes(x, !factimes(y, z)) 
    !facplus(!facplus(x, y), z) => !facplus(x, !facplus(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) 

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x1 + 2x1 + 3x0 + x0x2 >= 2 + x2 + 2x0x1 + 2x1 + 3x0 + x0x2 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[!factimes(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x0x2 + 3x2 + x1x2 >= 2 + x1 + 2x0 + 2x0x2 + 3x2 + x1x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] 
  [[!factimes(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + x0x1 + x0x1x2 + x0x2 + x1x2 >= x0 + x1 + x2 + x0x1 + x0x1x2 + x0x2 + x1x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] 
  [[!facplus(!facplus(_x0, _x1), _x2)]] = 6 + x2 + 2x1 + 4x0 > 4 + x2 + 2x0 + 2x1 = [[!facplus(_x0, !facplus(_x1, _x2))]] 
  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + F0(0) + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 4F0(2 + x1 + x2) > 4 + x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 4F0(2 + x1 + x2) > 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

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

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

The following interpretation satisfies the requirements:

  !facplus = \y0y1.2 + y1 + 2y0 
  !factimes = \y0y1.y0 + y1 + 2y0y1 
  filter = \G0y1.3 + 3y1 + G0(0) 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 2 + x2 + 2x0x2 + 2x1 + 4x0x1 + 5x0 >= 2 + x2 + 2x0x2 + 2x1 + 3x0 + 4x0x1 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[!factimes(!facplus(_x0, _x1), _x2)]] = 2 + x1 + 2x0 + 2x1x2 + 4x0x2 + 5x2 >= 2 + x1 + 2x0 + 2x1x2 + 3x2 + 4x0x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] 
  [[!factimes(!factimes(_x0, _x1), _x2)]] = x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 >= x0 + x1 + x2 + 2x0x1 + 2x0x2 + 2x1x2 + 4x0x1x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] 
  [[filter(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  filter(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)) 
  !factimes(!facplus(X, Y), Z) >? !facplus(!factimes(Z, X), !factimes(Z, Y)) 
  !factimes(!factimes(X, Y), Z) >? !factimes(X, !factimes(Y, 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 = \y0y1.1 + 2y0 + 2y0y1 + 2y1 

Using this interpretation, the requirements translate to:

  [[!factimes(_x0, !facplus(_x1, _x2))]] = 5 + 2x0x1 + 2x0x2 + 2x1 + 2x2 + 6x0 > 4 + 2x0x1 + 2x0x2 + 2x1 + 2x2 + 4x0 = [[!facplus(!factimes(_x0, _x1), !factimes(_x0, _x2))]] 
  [[!factimes(!facplus(_x0, _x1), _x2)]] = 5 + 2x0 + 2x0x2 + 2x1 + 2x1x2 + 6x2 > 4 + 2x0 + 2x0x2 + 2x1 + 2x1x2 + 4x2 = [[!facplus(!factimes(_x2, _x0), !factimes(_x2, _x1))]] 
  [[!factimes(!factimes(_x0, _x1), _x2)]] = 3 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 + 4x1 + 4x1x2 + 4x2 >= 3 + 4x0 + 4x0x1 + 4x0x1x2 + 4x0x2 + 4x1 + 4x1x2 + 4x2 = [[!factimes(_x0, !factimes(_x1, _x2))]] 

We can thus remove the following rules:

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[!factimes(!factimes(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[!factimes(_x0, !factimes(_x1, _x2))]] 

We can thus remove the following rules:

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