We consider the system AotoYamada_05__020.

  Alphabet:

    0 : [] --> a 
    comp : [b -> b * b -> b] --> b -> b 
    plus : [a * a] --> a 
    s : [a] --> a 
    times : [a * a] --> a 
    twice : [b -> b] --> b -> b 

  Rules:

    plus(0, x) => x 
    plus(s(x), y) => s(plus(x, y)) 
    times(0, x) => 0 
    times(s(x), y) => plus(times(x, y), y) 
    comp(f, g) x => f (g x) 
    twice(f) => comp(f, f) 

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)) 
  times(0, X) >? 0 
  times(s(X), Y) >? plus(times(X, Y), Y) 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

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

We choose Lex = {} and Mul = {0, @_{o -> o}, comp, plus, s, times, twice}, and the following precedence: 0 > times > plus > s > twice > comp > @_{o -> o}

With these choices, we have:

  1] plus(0, X) > X  because [2], by definition 
  2] plus*(0, X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

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

  11] times(0, X) >= 0  because [12], by (Star) 
  12] times*(0, X) >= 0  because [13], by (Select) 
  13] 0 >= 0  by (Fun) 

  14] times(s(X), Y) > plus(times(X, Y), Y)  because [15], by definition 
  15] times*(s(X), Y) >= plus(times(X, Y), Y)  because times > plus, [16] and [21], by (Copy) 
  16] times*(s(X), Y) >= times(X, Y)  because times in Mul, [17] and [20], by (Stat) 
  17] s(X) > X  because [18], by definition 
  18] s*(X) >= X  because [19], by (Select) 
  19] X >= X  by (Meta) 
  20] Y >= Y  by (Meta) 
  21] times*(s(X), Y) >= Y  because [20], by (Select) 

  22] @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [23], by (Star) 
  23] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [24], by (Select) 
  24] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [25] 
  25] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [26] and [28], by (Copy) 
  26] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [27], by (Select) 
  27] F >= F  by (Meta) 
  28] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [29] and [31], by (Copy) 
  29] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [30], by (Select) 
  30] G >= G  by (Meta) 
  31] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [32], by (Select) 
  32] @_{o -> o}*(comp(F, G), X) >= X  because [33], by (Select) 
  33] X >= X  by (Meta) 

  34] twice(F) >= comp(F, F)  because [35], by (Star) 
  35] twice*(F) >= comp(F, F)  because twice > comp, [36] and [36], by (Copy) 
  36] twice*(F) >= F  because [37], by (Select) 
  37] F >= F  by (Meta) 

We can thus remove the following rules:

  plus(0, X) => X 
  times(s(X), Y) => plus(times(X, Y), 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)) 
  times(0, X) >? 0 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

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

Argument functions:

  [[0]] = _|_ 
  [[s(x_1)]] = x_1 

We choose Lex = {} and Mul = {@_{o -> o}, comp, plus, times, twice}, and the following precedence: plus > times > twice > comp > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus(X, Y) >= plus(X, Y) 
  times(_|_, X) > _|_ 
  @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X)) 
  twice(F) >= comp(F, F) 

With these choices, we have:

  1] plus(X, Y) >= plus(X, Y)  because plus in Mul, [2] and [3], by (Fun) 
  2] X >= X  by (Meta) 
  3] Y >= Y  by (Meta) 

  4] times(_|_, X) > _|_  because [5], by definition 
  5] times*(_|_, X) >= _|_  by (Bot) 

  6] @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [7], by (Star) 
  7] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [8], by (Select) 
  8] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [9] 
  9] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [10] and [12], by (Copy) 
  10] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [11], by (Select) 
  11] F >= F  by (Meta) 
  12] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [13] and [15], by (Copy) 
  13] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [14], by (Select) 
  14] G >= G  by (Meta) 
  15] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [16], by (Select) 
  16] @_{o -> o}*(comp(F, G), X) >= X  because [17], by (Select) 
  17] comp(F, G) @_{o -> o}*(comp(F, G), X) >= X  because [18] 
  18] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [19], by (Select) 
  19] @_{o -> o}*(comp(F, G), X) >= X  because [20], by (Select) 
  20] X >= X  by (Meta) 

  21] twice(F) >= comp(F, F)  because [22], by (Star) 
  22] twice*(F) >= comp(F, F)  because twice > comp, [23] and [23], by (Copy) 
  23] twice*(F) >= F  because [24], by (Select) 
  24] F >= F  by (Meta) 

We can thus remove the following rules:

  times(0, X) => 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(s(X), Y) >? s(plus(X, Y)) 
  comp(F, G) X >? F (G X) 
  twice(F) >? comp(F, F) 

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

Argument functions:

  [[s(x_1)]] = x_1 

We choose Lex = {} and Mul = {@_{o -> o}, comp, plus, twice}, and the following precedence: plus > twice > comp > @_{o -> o}

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus(X, Y) >= plus(X, Y) 
  @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X)) 
  twice(F) > comp(F, F) 

With these choices, we have:

  1] plus(X, Y) >= plus(X, Y)  because plus in Mul, [2] and [3], by (Fun) 
  2] X >= X  by (Meta) 
  3] Y >= Y  by (Meta) 

  4] @_{o -> o}(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [5], by (Star) 
  5] @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [6], by (Select) 
  6] comp(F, G) @_{o -> o}*(comp(F, G), X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because [7] 
  7] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [8] and [10], by (Copy) 
  8] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= F  because [9], by (Select) 
  9] F >= F  by (Meta) 
  10] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [11] and [13], by (Copy) 
  11] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= G  because [12], by (Select) 
  12] G >= G  by (Meta) 
  13] comp*(F, G, @_{o -> o}*(comp(F, G), X)) >= X  because [14], by (Select) 
  14] @_{o -> o}*(comp(F, G), X) >= X  because [15], by (Select) 
  15] X >= X  by (Meta) 

  16] twice(F) > comp(F, F)  because [17], by definition 
  17] twice*(F) >= comp(F, F)  because twice > comp, [18] and [18], by (Copy) 
  18] twice*(F) >= F  because [19], by (Select) 
  19] F >= F  by (Meta) 

We can thus remove the following rules:

  twice(F) => comp(F, F) 

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)) 
  comp(F, G, X) >? F (G X) 

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

Argument functions:

  [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) 
  [[comp(x_1, x_2, x_3)]] = comp(x_1, x_3, x_2) 
  [[s(x_1)]] = x_1 

We choose Lex = {@_{o -> o}, comp} and Mul = {plus}, and the following precedence: comp > @_{o -> o} > plus

Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:

  plus(X, Y) >= plus(X, Y) 
  comp(F, G, X) > @_{o -> o}(F, @_{o -> o}(G, X)) 

With these choices, we have:

  1] plus(X, Y) >= plus(X, Y)  because plus in Mul, [2] and [3], by (Fun) 
  2] X >= X  by (Meta) 
  3] Y >= Y  by (Meta) 

  4] comp(F, G, X) > @_{o -> o}(F, @_{o -> o}(G, X))  because [5], by definition 
  5] comp*(F, G, X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because comp > @_{o -> o}, [6] and [8], by (Copy) 
  6] comp*(F, G, X) >= F  because [7], by (Select) 
  7] F >= F  by (Meta) 
  8] comp*(F, G, X) >= @_{o -> o}(G, X)  because comp > @_{o -> o}, [9] and [11], by (Copy) 
  9] comp*(F, G, X) >= G  because [10], by (Select) 
  10] G >= G  by (Meta) 
  11] comp*(F, G, X) >= X  because [12], by (Select) 
  12] X >= X  by (Meta) 

We can thus remove the following rules:

  comp(F, G, X) => F (G X) 

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.