/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
We consider the system theBenchmark.

  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).

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) 
  curry(F) X Y >? F X Y 
  add >? curry(plus) 

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

The following interpretation satisfies the requirements:

  0 = 3 
  add = \y0y1.3 + 3y0 + 3y1 
  curry = \G0y1y2.2G0(y1,y2) 
  plus = \y0y1.y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[plus 0 _x0]] = 6 + x0 > x0 = [[_x0]] 
  [[plus s(_x0) _x1]] = 6 + x1 + 2x0 > 3 + x1 + 2x0 = [[s(plus _x0 _x1)]] 
  [[curry(_F0) _x1 _x2]] = x1 + x2 + 2F0(x1,x2) >= x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] 
  [[add]] = \y0y1.3 + 3y0 + 3y1 > \y0y1.2y0 = [[curry(plus)]] 

We can thus remove the following rules:

  plus 0 X => X 
  plus s(X) Y => s(plus X Y) 
  add => curry(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]):

  curry(F, X, Y) >? F X Y 

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

The following interpretation satisfies the requirements:

  curry = \G0y1y2.1 + y1 + y2 + G0(y1,y2) 

Using this interpretation, the requirements translate to:

  [[curry(_F0, _x1, _x2)]] = 1 + x1 + x2 + F0(x1,x2) > x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] 

We can thus remove the following rules:

  curry(F, X, Y) => F 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.