We consider the system 06plusmult.

  Alphabet:

    mult : [N * N] --> N 
    plus : [N * N] --> N 
    s : [N] --> N 
    z : [] --> N 

  Rules:

    plus(z, x) => x 
    plus(s(x), y) => plus(x, s(y)) 
    plus(plus(x, y), u) => plus(x, plus(y, u)) 
    mult(z, x) => z 
    mult(s(x), y) => plus(mult(x, y), y) 
    mult(plus(x, y), u) => plus(mult(x, u), mult(y, u)) 

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(z, X) >? X 
  plus(s(X), Y) >? plus(X, s(Y)) 
  plus(plus(X, Y), Z) >? plus(X, plus(Y, Z)) 
  mult(z, X) >? z 
  mult(s(X), Y) >? plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) >? plus(mult(X, Z), mult(Y, Z)) 

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

The following interpretation satisfies the requirements:

  mult = \y0y1.y1 + 2y0y1 + 3y0 
  plus = \y0y1.1 + y0 + y1 
  s = \y0.3 + y0 
  z = 0 

Using this interpretation, the requirements translate to:

  [[plus(z, _x0)]] = 1 + x0 > x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 4 + x0 + x1 >= 4 + x0 + x1 = [[plus(_x0, s(_x1))]] 
  [[plus(plus(_x0, _x1), _x2)]] = 2 + x0 + x1 + x2 >= 2 + x0 + x1 + x2 = [[plus(_x0, plus(_x1, _x2))]] 
  [[mult(z, _x0)]] = x0 >= 0 = [[z]] 
  [[mult(s(_x0), _x1)]] = 9 + 2x0x1 + 3x0 + 7x1 > 1 + 2x0x1 + 2x1 + 3x0 = [[plus(mult(_x0, _x1), _x1)]] 
  [[mult(plus(_x0, _x1), _x2)]] = 3 + 2x0x2 + 2x1x2 + 3x0 + 3x1 + 3x2 > 1 + 2x0x2 + 2x1x2 + 2x2 + 3x0 + 3x1 = [[plus(mult(_x0, _x2), mult(_x1, _x2))]] 

We can thus remove the following rules:

  plus(z, X) => X 
  mult(s(X), Y) => plus(mult(X, Y), Y) 
  mult(plus(X, Y), Z) => plus(mult(X, Z), mult(Y, 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]):

  plus(s(X), Y) >? plus(X, s(Y)) 
  plus(plus(X, Y), Z) >? plus(X, plus(Y, Z)) 
  mult(z, X) >? z 

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

The following interpretation satisfies the requirements:

  mult = \y0y1.3 + y1 + 3y0 
  plus = \y0y1.3 + y1 + 3y0 
  s = \y0.3 + y0 
  z = 0 

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[plus(_x0, s(_x1))]] 
  [[plus(plus(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[plus(_x0, plus(_x1, _x2))]] 
  [[mult(z, _x0)]] = 3 + x0 > 0 = [[z]] 

We can thus remove the following rules:

  plus(s(X), Y) => plus(X, s(Y)) 
  plus(plus(X, Y), Z) => plus(X, plus(Y, Z)) 
  mult(z, X) => 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.