/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 
  0 : [] --> o 
  double : [o] --> o 
  s : [o] --> o 

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

  double(0) >? 0 
  double(s(X)) >? s(s(double(X))) 
  !plus(X, 0) >? X 
  !plus(X, s(Y)) >? s(!plus(X, Y)) 
  !plus(s(X), Y) >? s(!plus(X, Y)) 
  double(X) >? !plus(X, X) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + 2y1 
  0 = 0 
  double = \y0.3y0 
  s = \y0.2 + y0 

Using this interpretation, the requirements translate to:

  [[double(0)]] = 0 >= 0 = [[0]] 
  [[double(s(_x0))]] = 6 + 3x0 > 4 + 3x0 = [[s(s(double(_x0)))]] 
  [[!plus(_x0, 0)]] = x0 >= x0 = [[_x0]] 
  [[!plus(_x0, s(_x1))]] = 4 + x0 + 2x1 > 2 + x0 + 2x1 = [[s(!plus(_x0, _x1))]] 
  [[!plus(s(_x0), _x1)]] = 2 + x0 + 2x1 >= 2 + x0 + 2x1 = [[s(!plus(_x0, _x1))]] 
  [[double(_x0)]] = 3x0 >= 3x0 = [[!plus(_x0, _x0)]] 

We can thus remove the following rules:

  double(s(X)) => s(s(double(X))) 
  !plus(X, s(Y)) => s(!plus(X, 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]):

  double(0) >? 0 
  !plus(X, 0) >? X 
  !plus(s(X), Y) >? s(!plus(X, Y)) 
  double(X) >? !plus(X, X) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y1 + 2y0 
  0 = 0 
  double = \y0.3 + 3y0 
  s = \y0.1 + y0 

Using this interpretation, the requirements translate to:

  [[double(0)]] = 3 > 0 = [[0]] 
  [[!plus(_x0, 0)]] = 2x0 >= x0 = [[_x0]] 
  [[!plus(s(_x0), _x1)]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[s(!plus(_x0, _x1))]] 
  [[double(_x0)]] = 3 + 3x0 > 3x0 = [[!plus(_x0, _x0)]] 

We can thus remove the following rules:

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

  !plus(X, 0) >? X 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.3 + y0 + y1 
  0 = 3 

Using this interpretation, the requirements translate to:

  [[!plus(_x0, 0)]] = 6 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  !plus(X, 0) => 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.