/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES
We consider the system theBenchmark.

  Alphabet:

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

  Rules:

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

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

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

The following interpretation satisfies the requirements:

  f = \y0y1.3 + 3y0 + 3y1 
  f1 = \y0y1.y0 
  f2 = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[f(_x0)]] = \y0.3 + 3y0 + 3x0 > \y0.x0 = [[f1(_x0)]] 
  [[f1(_x0) _x1]] = x0 + x1 >= x0 + x1 = [[f2(_x0, _x1)]] 

We can thus remove the following rules:

  f(X) => f1(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]):

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

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

The following interpretation satisfies the requirements:

  f1 = \y0y1.3 + 3y0 + 3y1 
  f2 = \y0y1.y0 + y1 

Using this interpretation, the requirements translate to:

  [[f1(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f2(_x0, _x1)]] 

We can thus remove the following rules:

  f1(X, Y) => f2(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.