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

  f(0) => s(0) 
  f(s(0)) => s(s(0)) 
  f(s(0)) => !times(s(s(0)), f(0)) 
  f(!plus(X, s(0))) => !plus(s(s(0)), f(X)) 
  f(!plus(X, Y)) => !times(f(X), f(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]):

  f(0) >? s(0) 
  f(s(0)) >? s(s(0)) 
  f(s(0)) >? !times(s(s(0)), f(0)) 
  f(!plus(X, s(0))) >? !plus(s(s(0)), f(X)) 
  f(!plus(X, Y)) >? !times(f(X), f(Y)) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.3 + 3y0 + 3y1 
  !times = \y0y1.y0 + y1 
  0 = 3 
  f = \y0.3 + 3y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[f(0)]] = 12 > 6 = [[s(0)]] 
  [[f(s(0))]] = 21 > 9 = [[s(s(0))]] 
  [[f(s(0))]] = 21 >= 21 = [[!times(s(s(0)), f(0))]] 
  [[f(!plus(_x0, s(0)))]] = 66 + 9x0 > 39 + 9x0 = [[!plus(s(s(0)), f(_x0))]] 
  [[f(!plus(_x0, _x1))]] = 12 + 9x0 + 9x1 > 6 + 3x0 + 3x1 = [[!times(f(_x0), f(_x1))]] 

We can thus remove the following rules:

  f(0) => s(0) 
  f(s(0)) => s(s(0)) 
  f(!plus(X, s(0))) => !plus(s(s(0)), f(X)) 
  f(!plus(X, Y)) => !times(f(X), f(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]):

  f(s(0)) >? !times(s(s(0)), f(0)) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[f(s(0))]] = 3 > 2 = [[!times(s(s(0)), f(0))]] 

We can thus remove the following rules:

  f(s(0)) => !times(s(s(0)), f(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.