/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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.

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

  f(s(X), Y, Z, U, V) => f(X, Y, Z, U, V) 
  f(0, s(X), Y, Z, U) => f(X, X, Y, Z, U) 
  f(0, 0, s(X), Y, Z) => f(X, X, X, Y, Z) 
  f(0, 0, 0, s(X), Y) => f(X, X, X, X, Y) 
  f(0, 0, 0, 0, s(X)) => f(X, X, X, X, X) 
  f(0, 0, 0, 0, 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(s(X), Y, Z, U, V) >? f(X, Y, Z, U, V) 
  f(0, s(X), Y, Z, U) >? f(X, X, Y, Z, U) 
  f(0, 0, s(X), Y, Z) >? f(X, X, X, Y, Z) 
  f(0, 0, 0, s(X), Y) >? f(X, X, X, X, Y) 
  f(0, 0, 0, 0, s(X)) >? f(X, X, X, X, X) 
  f(0, 0, 0, 0, 0) >? 0 

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

The following interpretation satisfies the requirements:

  0 = 0 
  f = \y0y1y2y3y4.3 + y0 + y1 + y2 + 2y3 + 3y4 
  s = \y0.3 + 3y0 

Using this interpretation, the requirements translate to:

  [[f(s(_x0), _x1, _x2, _x3, _x4)]] = 6 + x1 + x2 + 2x3 + 3x0 + 3x4 > 3 + x0 + x1 + x2 + 2x3 + 3x4 = [[f(_x0, _x1, _x2, _x3, _x4)]] 
  [[f(0, s(_x0), _x1, _x2, _x3)]] = 6 + x1 + 2x2 + 3x0 + 3x3 > 3 + x1 + 2x0 + 2x2 + 3x3 = [[f(_x0, _x0, _x1, _x2, _x3)]] 
  [[f(0, 0, s(_x0), _x1, _x2)]] = 6 + 2x1 + 3x0 + 3x2 > 3 + 2x1 + 3x0 + 3x2 = [[f(_x0, _x0, _x0, _x1, _x2)]] 
  [[f(0, 0, 0, s(_x0), _x1)]] = 9 + 3x1 + 6x0 > 3 + 3x1 + 5x0 = [[f(_x0, _x0, _x0, _x0, _x1)]] 
  [[f(0, 0, 0, 0, s(_x0))]] = 12 + 9x0 > 3 + 8x0 = [[f(_x0, _x0, _x0, _x0, _x0)]] 
  [[f(0, 0, 0, 0, 0)]] = 3 > 0 = [[0]] 

We can thus remove the following rules:

  f(s(X), Y, Z, U, V) => f(X, Y, Z, U, V) 
  f(0, s(X), Y, Z, U) => f(X, X, Y, Z, U) 
  f(0, 0, s(X), Y, Z) => f(X, X, X, Y, Z) 
  f(0, 0, 0, s(X), Y) => f(X, X, X, X, Y) 
  f(0, 0, 0, 0, s(X)) => f(X, X, X, X, X) 
  f(0, 0, 0, 0, 0) => 0 

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.