We consider the system Applicative_first_order_05__#3.16.

  Alphabet:

    0 : [] --> 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 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    true : [] --> b 

  Rules:

    times(x, 0) => 0 
    times(x, s(y)) => plus(times(x, y), x) 
    plus(x, 0) => x 
    plus(0, x) => x 
    plus(x, s(y)) => s(plus(x, y)) 
    plus(s(x), y) => s(plus(x, y)) 
    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]):

  times(X, 0) >? 0 
  times(X, s(Y)) >? plus(times(X, Y), X) 
  plus(X, 0) >? X 
  plus(0, X) >? X 
  plus(X, s(Y)) >? s(plus(X, Y)) 
  plus(s(X), Y) >? s(plus(X, Y)) 
  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 = 0 
  cons = \y0y1.2 + y1 + 2y0 
  false = 3 
  filter = \G0y1.1 + 3y1 + G0(y1) + 2y1G0(y1) 
  filter2 = \y0G1y2y3.2y0 + 2y2 + 3y3 + G1(y3) + 2y3G1(y3) 
  map = \G0y1.2y1 + G0(y1) + y1G0(y1) 
  nil = 0 
  plus = \y0y1.y0 + y1 
  s = \y0.1 + y0 
  times = \y0y1.2y0 + 2y1 + y0y1 
  true = 3 

Using this interpretation, the requirements translate to:

  [[times(_x0, 0)]] = 2x0 >= 0 = [[0]] 
  [[times(_x0, s(_x1))]] = 2 + 2x1 + 3x0 + x0x1 > 2x1 + 3x0 + x0x1 = [[plus(times(_x0, _x1), _x0)]] 
  [[plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[plus(0, _x0)]] = x0 >= x0 = [[_x0]] 
  [[plus(_x0, s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 4 + 2x2 + 4x1 + 2x1F0(2 + x2 + 2x1) + 3F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 2 + 2x1 + 2x2 + F0(x2) + 2F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 7 + 3x2 + 6x1 + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 5F0(2 + x2 + 2x1) > 3x2 + 4x1 + F0(x2) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) > 3 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) > 1 + 3x2 + F0(x2) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  times(X, s(Y)) => plus(times(X, Y), X) 
  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]):

  times(X, 0) >? 0 
  plus(X, 0) >? X 
  plus(0, X) >? X 
  plus(X, s(Y)) >? s(plus(X, Y)) 
  plus(s(X), Y) >? s(plus(X, Y)) 
  map(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  0 = 0 
  map = \G0y1.3 + 3y1 + G0(0) 
  nil = 0 
  plus = \y0y1.3 + 3y0 + 3y1 
  s = \y0.y0 
  times = \y0y1.3 + y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[times(_x0, 0)]] = 3 + x0 > 0 = [[0]] 
  [[plus(_x0, 0)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[plus(0, _x0)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[plus(_x0, s(_x1))]] = 3 + 3x0 + 3x1 >= 3 + 3x0 + 3x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = 3 + 3x0 + 3x1 >= 3 + 3x0 + 3x1 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  times(X, 0) => 0 
  plus(X, 0) => X 
  plus(0, X) => X 
  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]):

  plus(X, s(Y)) >? s(plus(X, Y)) 
  plus(s(X), Y) >? s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  plus = \y0y1.3y0 + 3y1 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[plus(_x0, s(_x1))]] = 6 + 3x0 + 3x1 > 2 + 3x0 + 3x1 = [[s(plus(_x0, _x1))]] 
  [[plus(s(_x0), _x1)]] = 6 + 3x0 + 3x1 > 2 + 3x0 + 3x1 = [[s(plus(_x0, _x1))]] 

We can thus remove the following rules:

  plus(X, s(Y)) => s(plus(X, Y)) 
  plus(s(X), Y) => s(plus(X, Y)) 

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.