/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.

  !870 : [o * o] --> o 
  and : [o * o] --> o 
  false : [] --> o 
  if : [o * o * o] --> o 
  implies : [o * o] --> o 
  not : [o] --> o 
  or : [o * o] --> o 
  true : [] --> o 

  not(X) => if(X, false, true) 
  and(X, Y) => if(X, Y, false) 
  or(X, Y) => if(X, true, Y) 
  implies(X, Y) => if(X, Y, true) 
  !870(X, X) => true 
  !870(X, Y) => if(X, Y, not(Y)) 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  if(X, X, if(X, false, true)) => true 
  !870(X, Y) => if(X, Y, if(Y, false, true)) 

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(X) >? if(X, false, true) 
  and(X, Y) >? if(X, Y, false) 
  or(X, Y) >? if(X, true, Y) 
  implies(X, Y) >? if(X, Y, true) 
  !870(X, X) >? true 
  !870(X, Y) >? if(X, Y, not(Y)) 
  if(true, X, Y) >? X 
  if(false, X, Y) >? Y 
  if(X, X, if(X, false, true)) >? true 
  !870(X, Y) >? if(X, Y, if(Y, false, true)) 

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

The following interpretation satisfies the requirements:

  !870 = \y0y1.3 + 3y0 + 3y1 
  and = \y0y1.3 + 3y0 + 3y1 
  false = 1 
  if = \y0y1y2.y0 + y1 + y2 
  implies = \y0y1.3 + 3y0 + 3y1 
  not = \y0.3 + y0 
  or = \y0y1.3 + 3y0 + 3y1 
  true = 1 

Using this interpretation, the requirements translate to:

  [[not(_x0)]] = 3 + x0 > 2 + x0 = [[if(_x0, false, true)]] 
  [[and(_x0, _x1)]] = 3 + 3x0 + 3x1 > 1 + x0 + x1 = [[if(_x0, _x1, false)]] 
  [[or(_x0, _x1)]] = 3 + 3x0 + 3x1 > 1 + x0 + x1 = [[if(_x0, true, _x1)]] 
  [[implies(_x0, _x1)]] = 3 + 3x0 + 3x1 > 1 + x0 + x1 = [[if(_x0, _x1, true)]] 
  [[!870(_x0, _x0)]] = 3 + 6x0 > 1 = [[true]] 
  [[!870(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + 2x1 = [[if(_x0, _x1, not(_x1))]] 
  [[if(true, _x0, _x1)]] = 1 + x0 + x1 > x0 = [[_x0]] 
  [[if(false, _x0, _x1)]] = 1 + x0 + x1 > x1 = [[_x1]] 
  [[if(_x0, _x0, if(_x0, false, true))]] = 2 + 3x0 > 1 = [[true]] 
  [[!870(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x0 + 2x1 = [[if(_x0, _x1, if(_x1, false, true))]] 

We can thus remove the following rules:

  not(X) => if(X, false, true) 
  and(X, Y) => if(X, Y, false) 
  or(X, Y) => if(X, true, Y) 
  implies(X, Y) => if(X, Y, true) 
  !870(X, X) => true 
  if(true, X, Y) => X 
  if(false, X, Y) => Y 
  if(X, X, if(X, false, true)) => true 
  !870(X, Y) => if(X, Y, if(Y, false, true)) 

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

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

  !870(X, Y) >? if(X, Y, not(Y)) 

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

The following interpretation satisfies the requirements:

  !870 = \y0y1.3 + 3y0 + 3y1 
  if = \y0y1y2.y0 + y1 + y2 
  not = \y0.y0 

Using this interpretation, the requirements translate to:

  [[!870(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + 2x1 = [[if(_x0, _x1, not(_x1))]] 

We can thus remove the following rules:

  !870(X, Y) => if(X, Y, not(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.