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


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We consider the system theBenchmark.

  Alphabet:

    0 : [] --> o 
    either : [o * o] --> o 
    f : [o -> o * o * o] --> o 
    g : [o * o] --> o 
    s : [o] --> o 

  Rules:

    f(h, x, 0) => 0 
    f(h, x, s(y)) => g(y, either(y, h x)) 
    g(x, y) => f(/\z.s(0), y, x) 

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(F, X, 0) >? 0 
  f(F, X, s(Y)) >? g(Y, either(Y, F X)) 
  g(X, Y) >? f(/\x.s(0), Y, X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  either = \y0y1.y0 + y1 
  f = \G0y1y2.1 + y1 + 2y2 + G0(y1) 
  g = \y0y1.1 + y1 + 2y0 
  s = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[f(_F0, _x1, 0)]] = 1 + x1 + F0(x1) > 0 = [[0]] 
  [[f(_F0, _x1, s(_x2))]] = 1 + x1 + 4x2 + F0(x1) >= 1 + x1 + 3x2 + F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] 
  [[g(_x0, _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[f(/\x.s(0), _x1, _x0)]] 

We can thus remove the following rules:

  f(F, X, 0) => 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(F, X, s(Y)) >? g(Y, either(Y, F X)) 
  g(X, Y) >? f(/\x.s(0), Y, X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  either = \y0y1.y0 + y1 
  f = \G0y1y2.2y1 + 3y2 + 2G0(y1) 
  g = \y0y1.2 + 2y1 + 3y0 
  s = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(_F0, _x1, s(_x2))]] = 3 + 2x1 + 9x2 + 2F0(x1) > 2 + 2x1 + 5x2 + 2F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] 
  [[g(_x0, _x1)]] = 2 + 2x1 + 3x0 >= 2 + 2x1 + 3x0 = [[f(/\x.s(0), _x1, _x0)]] 

We can thus remove the following rules:

  f(F, X, s(Y)) => g(Y, either(Y, F 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]):

  g(X, Y) >? f(/\x.s(0), Y, X) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \G0y1y2.y1 + y2 + G0(0) 
  g = \y0y1.3 + 3y0 + 3y1 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[g(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f(/\x.s(0), _x1, _x0)]] 

We can thus remove the following rules:

  g(X, Y) => f(/\x.s(0), Y, 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.