/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

We are asked to determine termination of the following first-order TRS.

  0 : [] --> o 
  activate : [o] --> o 
  cons : [o * o] --> o 
  n!6220!6220zeros : [] --> o 
  tail : [o] --> o 
  zeros : [] --> o 

  zeros => cons(0, n!6220!6220zeros) 
  tail(cons(X, Y)) => activate(Y) 
  zeros => n!6220!6220zeros 
  activate(n!6220!6220zeros) => zeros 
  activate(X) => X 

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

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

  zeros >? cons(0, n!6220!6220zeros) 
  tail(cons(X, Y)) >? activate(Y) 
  zeros >? n!6220!6220zeros 
  activate(n!6220!6220zeros) >? zeros 
  activate(X) >? X 

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

The following interpretation satisfies the requirements:

  0 = 0 
  activate = \y0.3 + y0 
  cons = \y0y1.1 + y0 + y1 
  n!6220!6220zeros = 1 
  tail = \y0.3 + 3y0 
  zeros = 3 

Using this interpretation, the requirements translate to:

  [[zeros]] = 3 > 2 = [[cons(0, n!6220!6220zeros)]] 
  [[tail(cons(_x0, _x1))]] = 6 + 3x0 + 3x1 > 3 + x1 = [[activate(_x1)]] 
  [[zeros]] = 3 > 1 = [[n!6220!6220zeros]] 
  [[activate(n!6220!6220zeros)]] = 4 > 3 = [[zeros]] 
  [[activate(_x0)]] = 3 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  zeros => cons(0, n!6220!6220zeros) 
  tail(cons(X, Y)) => activate(Y) 
  zeros => n!6220!6220zeros 
  activate(n!6220!6220zeros) => zeros 
  activate(X) => X 

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.