We consider the system Applicative_05__Ex3Lists.

  Alphabet:

    append : [b * b] --> b 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    append(append(x, y), z) => append(x, append(y, z)) 
    map(f, append(x, y)) => append(map(f, x), map(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]):

  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  append(append(X, Y), Z) >? append(X, append(Y, Z)) 
  map(F, append(X, Y)) >? append(map(F, X), map(F, Y)) 

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

The following interpretation satisfies the requirements:

  append = Lam[y0;y1].3 + y1 + 2*y0 
  cons = Lam[y0;y1].3 + y0 + y1 
  map = Lam[G0;y1].2 + 3*y1 + G0(0) + 3*y1*G0(y1) 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[append(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 9 + x2 + 2*x0 + 2*x1 > 6 + x0 + x2 + 2*x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 11 + 3*x1 + 3*x2 + F0(0) + 3*x1*F0(3 + x1 + x2) + 3*x2*F0(3 + x1 + x2) + 9*F0(3 + x1 + x2) > 5 + x1 + 3*x2 + F0(0) + F0(x1) + 3*x2*F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = 9 + x2 + 2*x1 + 4*x0 > 6 + x2 + 2*x0 + 2*x1 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 11 + 3*x2 + 6*x1 + F0(0) + 3*x2*F0(3 + x2 + 2*x1) + 6*x1*F0(3 + x2 + 2*x1) + 9*F0(3 + x2 + 2*x1) > 9 + 3*x2 + 6*x1 + 3*x2*F0(x2) + 3*F0(0) + 6*x1*F0(x1) = [[append(map(_F0, _x1), map(_F0, _x2))]] 

We can thus remove the following rules:

  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  append(append(X, Y), Z) => append(X, append(Y, Z)) 
  map(F, append(X, Y)) => append(map(F, X), map(F, 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.