We consider the system AotoYamada_05__004.

  Alphabet:

    0 : [] --> b 
    cons : [a * b] --> b 
    nil : [] --> b 
    plus : [b * b] --> b 
    s : [b] --> b 
    sumwith : [a -> b * b] --> b 

  Rules:

    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    sumwith(f, nil) => nil 
    sumwith(f, cons(x, y)) => plus(f x, sumwith(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]):

  plus(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  sumwith(F, nil) >? nil 
  sumwith(F, cons(X, Y)) >? plus(F X, sumwith(F, Y)) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  cons = Lam[y0;y1].3 + 3*y0 + 3*y1 
  nil = 2 
  plus = Lam[y0;y1].y0 + y1 
  s = Lam[y0].y0 
  sumwith = Lam[G0;y1].3 + 3*y1 + G0(y1) + 3*y1*G0(y1) 

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[s(plus(_x0, _x1))]] 
  [[sumwith(_F0, nil)]] = 9 + 7*F0(2) > 2 = [[nil]] 
  [[sumwith(_F0, cons(_x1, _x2))]] = 12 + 9*x1 + 9*x2 + 9*x1*F0(3 + 3*x1 + 3*x2) + 9*x2*F0(3 + 3*x1 + 3*x2) + 10*F0(3 + 3*x1 + 3*x2) > 3 + x1 + 3*x2 + F0(x1) + F0(x2) + 3*x2*F0(x2) = [[plus(_F0 _x1, sumwith(_F0, _x2))]] 

We can thus remove the following rules:

  plus(0, X) => X 
  sumwith(F, nil) => nil 
  sumwith(F, cons(X, Y)) => plus(F X, sumwith(F, Y)) 

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

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

  plus(s(X), Y) >? s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  plus = Lam[y0;y1].2*y1 + 3*y0 
  s = Lam[y0].2 + y0 

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 6 + 2*x1 + 3*x0 > 2 + 2*x1 + 3*x0 = [[s(plus(_x0, _x1))]] 

We can thus remove the following rules:

  plus(s(X), Y) => s(plus(X, 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.