We consider the system 515.

  Alphabet:

    cons : [a * alist] --> alist 
    map : [a -> a * alist] --> alist 
    nil : [] --> alist 

  Rules:

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

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[map(/\x._x0(x), nil)]] = 3 + 2F0(0) > 0 = [[nil]] 
  [[map(/\x._x0(x), cons(_x1, _x2))]] = 12 + 3x1 + 3x2 + 5F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) > 6 + 3x2 + F0(x1) + 2F0(x2) + x2F0(x2) = [[cons(_x0(_x1), map(/\x._x0(x), _x2))]] 
  [[map(/\x._x0(x), map(/\y._x1(y), _x2))]] = 12 + 9x2 + 2F0(3 + 3x2 + 2F1(x2) + x2F1(x2))F1(x2) + 3x2F0(3 + 3x2 + 2F1(x2) + x2F1(x2)) + 3x2F1(x2) + 5F0(3 + 3x2 + 2F1(x2) + x2F1(x2)) + 6F1(x2) + x2F0(3 + 3x2 + 2F1(x2) + x2F1(x2))F1(x2) > 3 + 3x2 + 2F0(F1(x2)) + x2F0(F1(x2)) = [[map(/\x._x0(_x1(x)), _x2)]] 

We can thus remove the following rules:

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