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 use a recursive path ordering as defined in [Kop12, Chapter 5].

We choose Lex = {plus} and Mul = {mult, s, z}, and the following precedence: mult > plus > s > z

With these choices, we have:

  1] plus(z, X) > X  because [2], by definition 
  2] plus*(z, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] plus(s(X), Y) >= plus(X, s(Y))  because [5], by (Star) 
  5] plus*(s(X), Y) >= plus(X, s(Y))  because [6], [9] and [11], by (Stat) 
  6] s(X) > X  because [7], by definition 
  7] s*(X) >= X  because [8], by (Select) 
  8] X >= X  by (Meta) 
  9] plus*(s(X), Y) >= X  because [10], by (Select) 
  10] s(X) >= X  because [7], by (Star) 
  11] plus*(s(X), Y) >= s(Y)  because plus > s and [12], by (Copy) 
  12] plus*(s(X), Y) >= Y  because [13], by (Select) 
  13] Y >= Y  by (Meta) 

  14] plus(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [15], by (Star) 
  15] plus*(plus(X, Y), Z) >= plus(X, plus(Y, Z))  because [16], [19] and [21], by (Stat) 
  16] plus(X, Y) > X  because [17], by definition 
  17] plus*(X, Y) >= X  because [18], by (Select) 
  18] X >= X  by (Meta) 
  19] plus*(plus(X, Y), Z) >= X  because [20], by (Select) 
  20] plus(X, Y) >= X  because [17], by (Star) 
  21] plus*(plus(X, Y), Z) >= plus(Y, Z)  because [22], [25], [27], [22] and [29], by (Stat) 
  22] plus(X, Y) > Y  because [23], by definition 
  23] plus*(X, Y) >= Y  because [24], by (Select) 
  24] Y >= Y  by (Meta) 
  25] plus*(plus(X, Y), Z) >= Y  because [26], by (Select) 
  26] plus(X, Y) >= Y  because [23], by (Star) 
  27] plus*(plus(X, Y), Z) >= Z  because [28], by (Select) 
  28] Z >= Z  by (Meta) 
  29] Z >= Z  by (Meta) 

  30] mult(z, X) > z  because [31], by definition 
  31] mult*(z, X) >= z  because mult > z, by (Copy) 

  32] mult(s(X), Y) > plus(mult(X, Y), Y)  because [33], by definition 
  33] mult*(s(X), Y) >= plus(mult(X, Y), Y)  because mult > plus, [34] and [39], by (Copy) 
  34] mult*(s(X), Y) >= mult(X, Y)  because mult in Mul, [35] and [38], by (Stat) 
  35] s(X) > X  because [36], by definition 
  36] s*(X) >= X  because [37], by (Select) 
  37] X >= X  by (Meta) 
  38] Y >= Y  by (Meta) 
  39] mult*(s(X), Y) >= Y  because [38], by (Select) 

  40] mult(plus(X, Y), Z) > plus(mult(X, Z), mult(Y, Z))  because [41], by definition 
  41] mult*(plus(X, Y), Z) >= plus(mult(X, Z), mult(Y, Z))  because mult > plus, [42] and [47], by (Copy) 
  42] mult*(plus(X, Y), Z) >= mult(X, Z)  because mult in Mul, [43] and [46], by (Stat) 
  43] plus(X, Y) > X  because [44], by definition 
  44] plus*(X, Y) >= X  because [45], by (Select) 
  45] X >= X  by (Meta) 
  46] Z >= Z  by (Meta) 
  47] mult*(plus(X, Y), Z) >= mult(Y, Z)  because mult in Mul, [48] and [46], by (Stat) 
  48] plus(X, Y) > Y  because [49], by definition 
  49] plus*(X, Y) >= Y  because [50], by (Select) 
  50] Y >= Y  by (Meta) 

We can thus remove the following rules:

  plus(z, X) => X 
  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 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)) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 2 + x1 + 2*x0 >= 2 + x1 + 2*x0 = [[plus(_x0, s(_x1))]] 
  [[plus(plus(_x0, _x1), _x2)]] = 6 + x2 + 2*x1 + 4*x0 > 4 + x2 + 2*x0 + 2*x1 = [[plus(_x0, plus(_x1, _x2))]] 

We can thus remove the following rules:

  plus(plus(X, Y), Z) => plus(X, plus(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)) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[plus(s(_x0), _x1)]] = 4 + x1 + 2*x0 > 2 + x1 + 2*x0 = [[plus(_x0, s(_x1))]] 

We can thus remove the following rules:

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