We consider the system 450.

  Alphabet:

    a : [] --> o 
    b : [] --> o 
    f : [o] --> o 

  Rules:

    a => b 
    f(a) => f(b) 

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

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

  a >? b 
  f(a) >? f(b) 

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

The following interpretation satisfies the requirements:

  a = 3 
  b = 0 
  f = \y0.y0 

Using this interpretation, the requirements translate to:

  [[a]] = 3 > 0 = [[b]] 
  [[f(a)]] = 3 > 0 = [[f(b)]] 

We can thus remove the following rules:

  a => b 
  f(a) => f(b) 

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.