/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 
  false : [] --> o 
  not : [o] --> o 
  odd : [o] --> o 
  s : [o] --> o 
  true : [] --> o 

  not(true) => false 
  not(false) => true 
  odd(0) => false 
  odd(s(X)) => not(odd(X)) 
  !plus(X, 0) => X 
  !plus(X, s(Y)) => s(!plus(X, Y)) 
  !plus(s(X), 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]):

  not(true) >? false 
  not(false) >? true 
  odd(0) >? false 
  odd(s(X)) >? not(odd(X)) 
  !plus(X, 0) >? X 
  !plus(X, s(Y)) >? s(!plus(X, Y)) 
  !plus(s(X), Y) >? s(!plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y0 + 2y1 
  0 = 3 
  false = 0 
  not = \y0.2 + y0 
  odd = \y0.2y0 
  s = \y0.1 + y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[not(true)]] = 2 > 0 = [[false]] 
  [[not(false)]] = 2 > 0 = [[true]] 
  [[odd(0)]] = 6 > 0 = [[false]] 
  [[odd(s(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[not(odd(_x0))]] 
  [[!plus(_x0, 0)]] = 6 + x0 > x0 = [[_x0]] 
  [[!plus(_x0, s(_x1))]] = 2 + x0 + 2x1 > 1 + x0 + 2x1 = [[s(!plus(_x0, _x1))]] 
  [[!plus(s(_x0), _x1)]] = 1 + x0 + 2x1 >= 1 + x0 + 2x1 = [[s(!plus(_x0, _x1))]] 

We can thus remove the following rules:

  not(true) => false 
  not(false) => true 
  odd(0) => false 
  !plus(X, 0) => 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]):

  odd(s(X)) >? not(odd(X)) 
  !plus(s(X), Y) >? s(!plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  !plus = \y0y1.y1 + 3y0 
  not = \y0.y0 
  odd = \y0.y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[odd(s(_x0))]] = 3 + x0 > x0 = [[not(odd(_x0))]] 
  [[!plus(s(_x0), _x1)]] = 9 + x1 + 3x0 > 3 + x1 + 3x0 = [[s(!plus(_x0, _x1))]] 

We can thus remove the following rules:

  odd(s(X)) => not(odd(X)) 
  !plus(s(X), Y) => s(!plus(X, 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.