We consider the system Applicative_first_order_05__hydra.

  Alphabet:

    0 : [] --> a 
    cons : [c * c] --> c 
    copy : [a * c * c] --> c 
    f : [c] --> c 
    false : [] --> b 
    filter : [c -> b * c] --> c 
    filter2 : [b * c -> b * c * c] --> c 
    map : [c -> c * c] --> c 
    n : [] --> a 
    nil : [] --> c 
    s : [a] --> a 
    true : [] --> b 

  Rules:

    f(cons(nil, x)) => x 
    f(cons(f(cons(nil, x)), y)) => copy(n, x, y) 
    copy(0, x, y) => f(y) 
    copy(s(x), y, z) => copy(x, y, cons(f(y), z)) 
    map(g, nil) => nil 
    map(g, cons(x, y)) => cons(g x, map(g, y)) 
    filter(g, nil) => nil 
    filter(g, cons(x, y)) => filter2(g x, g, x, y) 
    filter2(true, g, x, y) => cons(x, filter(g, y)) 
    filter2(false, g, x, y) => filter(g, 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(cons(nil, X)) >? X 
  f(cons(f(cons(nil, X)), Y)) >? copy(n, X, Y) 
  copy(0, X, Y) >? f(Y) 
  copy(s(X), Y, Z) >? copy(X, Y, cons(f(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:

  0 = 3 
  cons = \y0y1.2 + y0 + y1 
  copy = \y0y1y2.y0 + y1 + y2 + y0y1 + y0y2 
  f = \y0.2y0 
  false = 3 
  filter = \G0y1.1 + 2y1 + G0(y1) + y1G0(y1) 
  filter2 = \y0G1y2y3.1 + y0 + y2 + 2y3 + G1(y3) + y3G1(y3) 
  map = \G0y1.2y1 + G0(y1) + 2y1G0(y1) + 2G0(0) 
  n = 0 
  nil = 0 
  s = \y0.3 + 3y0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[f(cons(nil, _x0))]] = 4 + 2x0 > x0 = [[_x0]] 
  [[f(cons(f(cons(nil, _x0)), _x1))]] = 12 + 2x1 + 4x0 > x0 + x1 = [[copy(n, _x0, _x1)]] 
  [[copy(0, _x0, _x1)]] = 3 + 4x0 + 4x1 > 2x1 = [[f(_x1)]] 
  [[copy(s(_x0), _x1, _x2)]] = 3 + 3x0 + 3x0x1 + 3x0x2 + 4x1 + 4x2 > 2 + x2 + 3x0 + 3x0x1 + 3x1 + x0x2 = [[copy(_x0, _x1, cons(f(_x1), _x2))]] 
  [[map(_F0, nil)]] = 3F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 2F0(0) + 5F0(2 + x1 + x2) > 2 + x1 + 2x2 + F0(x1) + F0(x2) + 2x2F0(x2) + 2F0(0) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 5 + 2x1 + 2x2 + 3F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > 1 + 2x1 + 2x2 + F0(x1) + F0(x2) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(x2) + x2F0(x2) > 3 + x1 + 2x2 + F0(x2) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 4 + x1 + 2x2 + F0(x2) + x2F0(x2) > 1 + 2x2 + F0(x2) + x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  f(cons(nil, X)) => X 
  f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) 
  copy(0, X, Y) => f(Y) 
  copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) 
  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]):

  map(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  map = \G0y1.3 + 3y1 + G0(0) 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  map(F, nil) => nil 

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.