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


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YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  a : [] --> o 
  a!6220!6220b : [] --> o 
  a!6220!6220f : [o * o * o] --> o 
  b : [] --> o 
  f : [o * o * o] --> o 
  mark : [o] --> o 

  a!6220!6220f(a, X, X) => a!6220!6220f(X, a!6220!6220b, b) 
  a!6220!6220b => a 
  mark(f(X, Y, Z)) => a!6220!6220f(X, mark(Y), Z) 
  mark(b) => a!6220!6220b 
  mark(a) => a 
  a!6220!6220f(X, Y, Z) => f(X, Y, Z) 
  a!6220!6220b => b 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  a!6220!6220f(a, X, X) >? a!6220!6220f(X, a!6220!6220b, b) 
  a!6220!6220b >? a 
  mark(f(X, Y, Z)) >? a!6220!6220f(X, mark(Y), Z) 
  mark(b) >? a!6220!6220b 
  mark(a) >? a 
  a!6220!6220f(X, Y, Z) >? f(X, Y, Z) 
  a!6220!6220b >? b 

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

The following interpretation satisfies the requirements:

  a = 2 
  a!6220!6220b = 3 
  a!6220!6220f = \y0y1y2.3 + y1 + 2y0 + 2y2 
  b = 0 
  f = \y0y1y2.3 + y0 + y1 + 2y2 
  mark = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[a!6220!6220f(a, _x0, _x0)]] = 7 + 3x0 > 6 + 2x0 = [[a!6220!6220f(_x0, a!6220!6220b, b)]] 
  [[a!6220!6220b]] = 3 > 2 = [[a]] 
  [[mark(f(_x0, _x1, _x2))]] = 9 + 2x0 + 2x1 + 4x2 > 6 + 2x0 + 2x1 + 2x2 = [[a!6220!6220f(_x0, mark(_x1), _x2)]] 
  [[mark(b)]] = 3 >= 3 = [[a!6220!6220b]] 
  [[mark(a)]] = 7 > 2 = [[a]] 
  [[a!6220!6220f(_x0, _x1, _x2)]] = 3 + x1 + 2x0 + 2x2 >= 3 + x0 + x1 + 2x2 = [[f(_x0, _x1, _x2)]] 
  [[a!6220!6220b]] = 3 > 0 = [[b]] 

We can thus remove the following rules:

  a!6220!6220f(a, X, X) => a!6220!6220f(X, a!6220!6220b, b) 
  a!6220!6220b => a 
  mark(f(X, Y, Z)) => a!6220!6220f(X, mark(Y), Z) 
  mark(a) => a 
  a!6220!6220b => b 

We use rule removal, following [Kop12, Theorem 2.23].

This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):

  mark(b) >? a!6220!6220b 
  a!6220!6220f(X, Y, Z) >? f(X, Y, Z) 

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

The following interpretation satisfies the requirements:

  a!6220!6220b = 0 
  a!6220!6220f = \y0y1y2.3 + y0 + y1 + y2 
  b = 3 
  f = \y0y1y2.y0 + y1 + y2 
  mark = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[mark(b)]] = 6 > 0 = [[a!6220!6220b]] 
  [[a!6220!6220f(_x0, _x1, _x2)]] = 3 + x0 + x1 + x2 > x0 + x1 + x2 = [[f(_x0, _x1, _x2)]] 

We can thus remove the following rules:

  mark(b) => a!6220!6220b 
  a!6220!6220f(X, Y, Z) => f(X, Y, Z) 

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.