We consider the system 430.

  Alphabet:

    append : [natlist * natlist] --> natlist 
    cons : [nat * natlist] --> natlist 
    map : [nat -> nat * natlist] --> natlist 
    nil : [] --> natlist 

  Rules:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    append(append(X, Y), Z) => append(X, append(Y, Z)) 
    map(/\x.X(x), nil) => nil 
    map(/\x.X(x), cons(Y, Z)) => cons(X(Y), map(/\y.X(y), Z)) 

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

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

  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  append(append(X, Y), Z) >? append(X, append(Y, Z)) 
  map(/\x.X(x), nil) >? nil 
  map(/\x.X(x), cons(Y, Z)) >? cons(X(Y), map(/\y.X(y), Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y1 + 2y0 
  cons = \y0y1.1 + y0 + y1 
  map = \G0y1.3 + 3y1 + G0(0) + G0(y1) + 2y1G0(y1) 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x2 + 2x0 + 2x1 > 1 + x0 + x2 + 2x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = x2 + 2x1 + 4x0 >= x2 + 2x0 + 2x1 = [[append(_x0, append(_x1, _x2))]] 
  [[map(/\x._x0(x), nil)]] = 3 + 2F0(0) > 0 = [[nil]] 
  [[map(/\x._x0(x), cons(_x1, _x2))]] = 6 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > 4 + 3x2 + F0(0) + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_x0(_x1), map(/\x._x0(x), _x2))]] 

We can thus remove the following rules:

  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
  map(/\x.X(x), nil) => nil 
  map(/\x.X(x), cons(Y, Z)) => cons(X(Y), map(/\y.X(y), Z)) 

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

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

  append(nil, X) >? X 
  append(append(X, Y), Z) >? append(X, append(Y, Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.3 + y1 + 3y0 
  nil = 3 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  append(nil, X) => X 
  append(append(X, Y), Z) => append(X, append(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.