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

Argument functions:

  [[s(x_1)]] = x_1 

We choose Lex = {} and Mul = {0, @_{o -> o}, cons, nil, plus, sumwith}, and the following precedence: cons > 0 > sumwith > plus > nil > @_{o -> o}

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

  plus(0, X) > X 
  plus(X, Y) >= plus(X, Y) 
  sumwith(F, nil) > nil 
  sumwith(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y)) 

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(X, Y) >= plus(X, Y)  because plus in Mul, [5] and [6], by (Fun) 
  5] X >= X  by (Meta) 
  6] Y >= Y  by (Meta) 

  7] sumwith(F, nil) > nil  because [8], by definition 
  8] sumwith*(F, nil) >= nil  because sumwith > nil, by (Copy) 

  9] sumwith(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y))  because [10], by (Star) 
  10] sumwith*(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y))  because sumwith > plus, [11] and [18], by (Copy) 
  11] sumwith*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because sumwith > @_{o -> o}, [12] and [14], by (Copy) 
  12] sumwith*(F, cons(X, Y)) >= F  because [13], by (Select) 
  13] F >= F  by (Meta) 
  14] sumwith*(F, cons(X, Y)) >= X  because [15], by (Select) 
  15] cons(X, Y) >= X  because [16], by (Star) 
  16] cons*(X, Y) >= X  because [17], by (Select) 
  17] X >= X  by (Meta) 
  18] sumwith*(F, cons(X, Y)) >= sumwith(F, Y)  because sumwith in Mul, [19] and [20], by (Stat) 
  19] F >= F  by (Meta) 
  20] cons(X, Y) > Y  because [21], by definition 
  21] cons*(X, Y) >= Y  because [22], by (Select) 
  22] Y >= Y  by (Meta) 

We can thus remove the following rules:

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

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)) 
  sumwith(F, cons(X, Y)) >? plus(F X, sumwith(F, Y)) 

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

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

With these choices, we have:

  1] plus(s(X), Y) > s(plus(X, Y))  because [2], by definition 
  2] plus*(s(X), Y) >= s(plus(X, Y))  because plus > s and [3], by (Copy) 
  3] plus*(s(X), Y) >= plus(X, Y)  because plus in Mul, [4] and [7], by (Stat) 
  4] s(X) > X  because [5], by definition 
  5] s*(X) >= X  because [6], by (Select) 
  6] X >= X  by (Meta) 
  7] Y >= Y  by (Meta) 

  8] sumwith(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y))  because [9], by (Star) 
  9] sumwith*(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y))  because sumwith > plus, [10] and [17], by (Copy) 
  10] sumwith*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because sumwith > @_{o -> o}, [11] and [13], by (Copy) 
  11] sumwith*(F, cons(X, Y)) >= F  because [12], by (Select) 
  12] F >= F  by (Meta) 
  13] sumwith*(F, cons(X, Y)) >= X  because [14], by (Select) 
  14] cons(X, Y) >= X  because [15], by (Star) 
  15] cons*(X, Y) >= X  because [16], by (Select) 
  16] X >= X  by (Meta) 
  17] sumwith*(F, cons(X, Y)) >= sumwith(F, Y)  because sumwith in Mul, [18] and [19], by (Stat) 
  18] F >= F  by (Meta) 
  19] cons(X, Y) > Y  because [20], by definition 
  20] cons*(X, Y) >= Y  because [21], by (Select) 
  21] Y >= Y  by (Meta) 

We can thus remove the following rules:

  plus(s(X), Y) => s(plus(X, 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]):

  sumwith(F, cons(X, Y)) >? plus(F X, sumwith(F, Y)) 

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

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

With these choices, we have:

  1] sumwith(F, cons(X, Y)) > plus(@_{o -> o}(F, X), sumwith(F, Y))  because [2], by definition 
  2] sumwith*(F, cons(X, Y)) >= plus(@_{o -> o}(F, X), sumwith(F, Y))  because sumwith > plus, [3] and [10], by (Copy) 
  3] sumwith*(F, cons(X, Y)) >= @_{o -> o}(F, X)  because sumwith > @_{o -> o}, [4] and [6], by (Copy) 
  4] sumwith*(F, cons(X, Y)) >= F  because [5], by (Select) 
  5] F >= F  by (Meta) 
  6] sumwith*(F, cons(X, Y)) >= X  because [7], by (Select) 
  7] cons(X, Y) >= X  because [8], by (Star) 
  8] cons*(X, Y) >= X  because [9], by (Select) 
  9] X >= X  by (Meta) 
  10] sumwith*(F, cons(X, Y)) >= sumwith(F, Y)  because sumwith in Mul, [11] and [12], by (Stat) 
  11] F >= F  by (Meta) 
  12] cons(X, Y) > Y  because [13], by definition 
  13] cons*(X, Y) >= Y  because [14], by (Select) 
  14] Y >= Y  by (Meta) 

We can thus remove the following rules:

  sumwith(F, cons(X, Y)) => plus(F X, sumwith(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.