/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.

  !times : [o * o] --> o 
  0 : [] --> o 
  1 : [] --> o 
  i : [o] --> o 

  !times(i(X), X) => 1 
  !times(1, X) => X 
  !times(X, 0) => 0 
  !times(!times(X, Y), Z) => !times(X, !times(Y, Z)) 

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

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

  !times(i(X), X) >? 1 
  !times(1, X) >? X 
  !times(X, 0) >? 0 
  !times(!times(X, Y), Z) >? !times(X, !times(Y, Z)) 

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

The following interpretation satisfies the requirements:

  !times = \y0y1.3 + y1 + 3y0 
  0 = 0 
  1 = 0 
  i = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[!times(i(_x0), _x0)]] = 12 + 4x0 > 0 = [[1]] 
  [[!times(1, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[!times(_x0, 0)]] = 3 + 3x0 > 0 = [[0]] 
  [[!times(!times(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[!times(_x0, !times(_x1, _x2))]] 

We can thus remove the following rules:

  !times(i(X), X) => 1 
  !times(1, X) => X 
  !times(X, 0) => 0 
  !times(!times(X, Y), Z) => !times(X, !times(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.