/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.

  !dot : [o * o] --> o 
  1 : [] --> o 
  i : [o] --> o 

  !dot(1, X) => X 
  !dot(X, 1) => X 
  !dot(i(X), X) => 1 
  !dot(X, i(X)) => 1 
  i(1) => 1 
  i(i(X)) => X 
  !dot(i(X), !dot(X, Y)) => Y 
  !dot(X, !dot(i(X), Y)) => Y 

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

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

  !dot(1, X) >? X 
  !dot(X, 1) >? X 
  !dot(i(X), X) >? 1 
  !dot(X, i(X)) >? 1 
  i(1) >? 1 
  i(i(X)) >? X 
  !dot(i(X), !dot(X, Y)) >? Y 
  !dot(X, !dot(i(X), Y)) >? Y 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[!dot(1, _x0)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[!dot(_x0, 1)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[!dot(i(_x0), _x0)]] = 12 + 12x0 > 0 = [[1]] 
  [[!dot(_x0, i(_x0))]] = 12 + 12x0 > 0 = [[1]] 
  [[i(1)]] = 3 > 0 = [[1]] 
  [[i(i(_x0))]] = 12 + 9x0 > x0 = [[_x0]] 
  [[!dot(i(_x0), !dot(_x0, _x1))]] = 21 + 9x1 + 18x0 > x1 = [[_x1]] 
  [[!dot(_x0, !dot(i(_x0), _x1))]] = 39 + 9x1 + 30x0 > x1 = [[_x1]] 

We can thus remove the following rules:

  !dot(1, X) => X 
  !dot(X, 1) => X 
  !dot(i(X), X) => 1 
  !dot(X, i(X)) => 1 
  i(1) => 1 
  i(i(X)) => X 
  !dot(i(X), !dot(X, Y)) => Y 
  !dot(X, !dot(i(X), Y)) => Y 

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.