/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 
  f : [o] --> o 

  !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) 
  !plus(f(X), f(Y)) => f(!plus(X, Y)) 
  !plus(f(X), !plus(f(Y), Z)) => !plus(f(!plus(X, 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]):

  !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) 
  !plus(f(X), f(Y)) >? f(!plus(X, Y)) 
  !plus(f(X), !plus(f(Y), Z)) >? !plus(f(!plus(X, Y)), Z) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + y1 
  f = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[!plus(!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus(_x0, !plus(_x1, _x2))]] 
  [[!plus(f(_x0), f(_x1))]] = 2 + 2x0 + 2x1 > 1 + 2x0 + 2x1 = [[f(!plus(_x0, _x1))]] 
  [[!plus(f(_x0), !plus(f(_x1), _x2))]] = 2 + x2 + 2x0 + 2x1 > 1 + x2 + 2x0 + 2x1 = [[!plus(f(!plus(_x0, _x1)), _x2)]] 

We can thus remove the following rules:

  !plus(f(X), f(Y)) => f(!plus(X, Y)) 
  !plus(f(X), !plus(f(Y), Z)) => !plus(f(!plus(X, 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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  !plus(!plus(X, Y), Z) => !plus(X, !plus(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.