/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.

  !plus : [o * o] --> o 
  !times : [o * o] --> o 

  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 
  !plus(!times(X, Y), !plus(X, Z)) => !times(X, !plus(Y, Z)) 
  !plus(!times(X, Y), !plus(!times(X, Z), U)) => !plus(!times(X, !plus(Y, Z)), U) 

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(X, !plus(Y, Z)) >? !plus(!plus(X, Y), Z) 
  !plus(!times(X, Y), !plus(X, Z)) >? !times(X, !plus(Y, Z)) 
  !plus(!times(X, Y), !plus(!times(X, Z), U)) >? !plus(!times(X, !plus(Y, Z)), U) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.3 + y0 + 3y1 
  !times = \y0y1.y0 + 2y1 

Using this interpretation, the requirements translate to:

  [[!plus(_x0, !plus(_x1, _x2))]] = 12 + x0 + 3x1 + 9x2 > 6 + x0 + 3x1 + 3x2 = [[!plus(!plus(_x0, _x1), _x2)]] 
  [[!plus(!times(_x0, _x1), !plus(_x0, _x2))]] = 12 + 2x1 + 4x0 + 9x2 > 6 + x0 + 2x1 + 6x2 = [[!times(_x0, !plus(_x1, _x2))]] 
  [[!plus(!times(_x0, _x1), !plus(!times(_x0, _x2), _x3))]] = 12 + 2x1 + 4x0 + 6x2 + 9x3 > 9 + x0 + 2x1 + 3x3 + 6x2 = [[!plus(!times(_x0, !plus(_x1, _x2)), _x3)]] 

We can thus remove the following rules:

  !plus(X, !plus(Y, Z)) => !plus(!plus(X, Y), Z) 
  !plus(!times(X, Y), !plus(X, Z)) => !times(X, !plus(Y, Z)) 
  !plus(!times(X, Y), !plus(!times(X, Z), U)) => !plus(!times(X, !plus(Y, Z)), U) 

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.