We consider the system AotoYamada_05__005.

  Alphabet:

    0 : [] --> a 
    add : [] --> a -> a -> a 
    curry : [a -> a -> a] --> a -> a -> a 
    plus : [] --> a -> a -> a 
    s : [a] --> a 

  Rules:

    plus 0 x => x 
    plus s(x) y => s(plus x y) 
    curry(f) x y => f x y 
    add => curry(plus) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

Symbol curry is an encoding for application that is only used in innocuous ways.  We can simplify the program (without losing non-termination) by removing it.  This gives:

  Alphabet:

    0 : [] --> a 
    add : [] --> a -> a -> a 
    plus : [] --> a -> a -> a 
    s : [a] --> a 

  Rules:

    plus 0 X => X 
    plus s(X) Y => s(plus X Y) 
    add => plus 

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) 
  add >? plus 

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 -> o}, @_{o -> o}, add, plus}, and the following precedence: add = plus > 0 > @_{o -> o -> o} > @_{o -> o}

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

  @_{o -> o}(@_{o -> o -> o}(plus, 0), X) > X 
  @_{o -> o}(@_{o -> o -> o}(plus, X), Y) >= @_{o -> o}(@_{o -> o -> o}(plus, X), Y) 
  add >= plus 

With these choices, we have:

  1] @_{o -> o}(@_{o -> o -> o}(plus, 0), X) > X  because [2], by definition 
  2] @_{o -> o}*(@_{o -> o -> o}(plus, 0), X) >= X  because [3], by (Select) 
  3] X >= X  by (Meta) 

  4] @_{o -> o}(@_{o -> o -> o}(plus, X), Y) >= @_{o -> o}(@_{o -> o -> o}(plus, X), Y)  because @_{o -> o} in Mul, [5] and [8], by (Fun) 
  5] @_{o -> o -> o}(plus, X) >= @_{o -> o -> o}(plus, X)  because @_{o -> o -> o} in Mul, [6] and [7], by (Fun) 
  6] plus >= plus  by (Fun) 
  7] X >= X  by (Meta) 
  8] Y >= Y  by (Meta) 

  9] add >= plus  because add = plus, by (Fun) 

We can thus remove the following rules:

  plus 0 X => 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) 
  add >? plus 

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

Argument functions:

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

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

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

  @_{o -> o}(@_{o -> o -> o}(_|_, X), Y) >= @_{o -> o}(@_{o -> o -> o}(_|_, X), Y) 
  add > _|_ 

With these choices, we have:

  1] @_{o -> o}(@_{o -> o -> o}(_|_, X), Y) >= @_{o -> o}(@_{o -> o -> o}(_|_, X), Y)  because @_{o -> o} in Mul, [2] and [5], by (Fun) 
  2] @_{o -> o -> o}(_|_, X) >= @_{o -> o -> o}(_|_, X)  because @_{o -> o -> o} in Mul, [3] and [4], by (Fun) 
  3] _|_ >= _|_  by (Bot) 
  4] X >= X  by (Meta) 
  5] Y >= Y  by (Meta) 

  6] add > _|_  because [7], by definition 
  7] add* >= _|_  by (Bot) 

We can thus remove the following rules:

  add => plus 

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

We choose Lex = {} and Mul = {plus, s}, and the following precedence: 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) 

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.