/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.

  and : [o * o] --> o 
  false : [] --> o 
  implies : [o * o] --> o 
  not : [o] --> o 

  and(X, false) => false 
  and(X, not(false)) => X 
  not(not(X)) => X 
  implies(false, X) => not(false) 
  implies(X, false) => not(X) 
  implies(not(X), not(Y)) => implies(Y, and(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]):

  and(X, false) >? false 
  and(X, not(false)) >? X 
  not(not(X)) >? X 
  implies(false, X) >? not(false) 
  implies(X, false) >? not(X) 
  implies(not(X), not(Y)) >? implies(Y, and(X, Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.y0 + y1 
  false = 0 
  implies = \y0y1.3 + 2y0 + 2y1 
  not = \y0.1 + 2y0 

Using this interpretation, the requirements translate to:

  [[and(_x0, false)]] = x0 >= 0 = [[false]] 
  [[and(_x0, not(false))]] = 1 + x0 > x0 = [[_x0]] 
  [[not(not(_x0))]] = 3 + 4x0 > x0 = [[_x0]] 
  [[implies(false, _x0)]] = 3 + 2x0 > 1 = [[not(false)]] 
  [[implies(_x0, false)]] = 3 + 2x0 > 1 + 2x0 = [[not(_x0)]] 
  [[implies(not(_x0), not(_x1))]] = 7 + 4x0 + 4x1 > 3 + 2x0 + 4x1 = [[implies(_x1, and(_x0, _x1))]] 

We can thus remove the following rules:

  and(X, not(false)) => X 
  not(not(X)) => X 
  implies(false, X) => not(false) 
  implies(X, false) => not(X) 
  implies(not(X), not(Y)) => implies(Y, and(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]):

  and(X, false) >? false 

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

The following interpretation satisfies the requirements:

  and = \y0y1.3 + y0 + 3y1 
  false = 0 

Using this interpretation, the requirements translate to:

  [[and(_x0, false)]] = 3 + x0 > 0 = [[false]] 

We can thus remove the following rules:

  and(X, false) => false 

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.