We consider the system curry1.

  Alphabet:

    f : [] --> a -> b -> c 
    f1 : [a] --> b -> c 
    f2 : [a * b] --> c 

  Rules:

    f1(x) => f x 
    f2(x, y) => f1(x) 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]):

  f1(X) >? f(X) 
  f2(X, Y) >? f1(X) Y 

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

We choose Lex = {} and Mul = {@_{o -> o}, f, f1, f2}, and the following precedence: f1 = f2 > @_{o -> o} > f

With these choices, we have:

  1] f1(X) >= f(X)  because [2], by (Star) 
  2] f1*(X) >= f(X)  because f1 > f and [3], by (Copy) 
  3] f1*(X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

  5] f2(X, Y) > @_{o -> o}(f1(X), Y)  because [6], by definition 
  6] f2*(X, Y) >= @_{o -> o}(f1(X), Y)  because f2 > @_{o -> o}, [7] and [9], by (Copy) 
  7] f2*(X, Y) >= f1(X)  because f2 = f1, f2 in Mul and [8], by (Stat) 
  8] X >= X  by (Meta) 
  9] f2*(X, Y) >= Y  because [10], by (Select) 
  10] Y >= Y  by (Meta) 

We can thus remove the following rules:

  f2(X, Y) => f1(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]):

  f1(X) >? f(X) 

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

We choose Lex = {} and Mul = {f, f1}, and the following precedence: f1 > f

With these choices, we have:

  1] f1(X) > f(X)  because [2], by definition 
  2] f1*(X) >= f(X)  because f1 > f and [3], by (Copy) 
  3] f1*(X) >= X  because [4], by (Select) 
  4] X >= X  by (Meta) 

We can thus remove the following rules:

  f1(X) => f(X) 

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.