/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.

  and : [o * o] --> o 
  not : [o] --> o 
  or : [o * o] --> o 

  or(X, X) => X 
  and(X, X) => X 
  not(not(X)) => X 
  not(and(X, Y)) => or(not(X), not(Y)) 
  not(or(X, Y)) => and(not(X), not(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]):

  or(X, X) >? X 
  and(X, X) >? X 
  not(not(X)) >? X 
  not(and(X, Y)) >? or(not(X), not(Y)) 
  not(or(X, Y)) >? and(not(X), not(Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.1 + y0 + y1 
  not = \y0.y0 
  or = \y0y1.1 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[or(_x0, _x0)]] = 1 + 2x0 > x0 = [[_x0]] 
  [[and(_x0, _x0)]] = 1 + 2x0 > x0 = [[_x0]] 
  [[not(not(_x0))]] = x0 >= x0 = [[_x0]] 
  [[not(and(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[or(not(_x0), not(_x1))]] 
  [[not(or(_x0, _x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[and(not(_x0), not(_x1))]] 

We can thus remove the following rules:

  or(X, X) => X 
  and(X, 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]):

  not(not(X)) >? X 
  not(and(X, Y)) >? or(not(X), not(Y)) 
  not(or(X, Y)) >? and(not(X), not(Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.3 + y0 + y1 
  not = \y0.3y0 
  or = \y0y1.3 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[not(not(_x0))]] = 9x0 >= x0 = [[_x0]] 
  [[not(and(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + 3x0 + 3x1 = [[or(not(_x0), not(_x1))]] 
  [[not(or(_x0, _x1))]] = 9 + 3x0 + 3x1 > 3 + 3x0 + 3x1 = [[and(not(_x0), not(_x1))]] 

We can thus remove the following rules:

  not(and(X, Y)) => or(not(X), not(Y)) 
  not(or(X, Y)) => and(not(X), not(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(not(X)) >? X 

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

The following interpretation satisfies the requirements:

  not = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[not(not(_x0))]] = 6 + x0 > x0 = [[_x0]] 

We can thus remove the following rules:

  not(not(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.