We consider the system merge.

  Alphabet:

    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    merge : [list * list * list] --> list 
    nil : [] --> list 

  Rules:

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

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

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

The following interpretation satisfies the requirements:

  cons = Lam[y0;y1].3 + y0 + y1 
  map = Lam[G0;y1].2 + 3*y1 + G0(0) + 3*y1*G0(y1) + 3*G0(y1) 
  merge = Lam[y0;y1;y2].3 + 2*y2 + 3*y0 + 3*y1 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[merge(nil, nil, _x0)]] = 3 + 2*x0 > x0 = [[_x0]] 
  [[merge(nil, cons(_x0, _x1), _x2)]] = 12 + 2*x2 + 3*x0 + 3*x1 > 9 + 2*x0 + 2*x2 + 3*x1 = [[merge(_x1, nil, cons(_x0, _x2))]] 
  [[merge(cons(_x0, _x1), _x2, _x3)]] = 12 + 2*x3 + 3*x0 + 3*x1 + 3*x2 > 9 + 2*x0 + 2*x3 + 3*x1 + 3*x2 = [[merge(_x2, _x1, cons(_x0, _x3))]] 
  [[map(_F0, nil)]] = 2 + 4*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) + 12*F0(3 + x1 + x2) > 5 + x1 + 3*x2 + F0(0) + F0(x1) + 3*x2*F0(x2) + 3*F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 

We can thus remove the following rules:

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