/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES
We consider the system theBenchmark.

  Alphabet:

    and : [c * c] --> c 
    cons : [a * b] --> b 
    false : [] --> c 
    forall : [a -> c * b] --> c 
    forsome : [a -> c * b] --> c 
    nil : [] --> b 
    or : [c * c] --> c 
    true : [] --> c 

  Rules:

    and(true, true) => true 
    and(x, false) => false 
    and(false, x) => false 
    or(true, x) => true 
    or(x, true) => true 
    or(false, false) => false 
    forall(f, nil) => true 
    forall(f, cons(x, y)) => and(f x, forall(f, y)) 
    forsome(f, nil) => false 
    forsome(f, cons(x, y)) => or(f x, forsome(f, y)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

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(true, true) >? true 
  and(X, false) >? false 
  and(false, X) >? false 
  or(true, X) >? true 
  or(X, true) >? true 
  or(false, false) >? false 
  forall(F, nil) >? true 
  forall(F, cons(X, Y)) >? and(F X, forall(F, Y)) 
  forsome(F, nil) >? false 
  forsome(F, cons(X, Y)) >? or(F X, forsome(F, Y)) 

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

The following interpretation satisfies the requirements:

  and = \y0y1.1 + y0 + y1 
  cons = \y0y1.3 + 3y0 + 3y1 
  false = 0 
  forall = \G0y1.3y1 + 2G0(y1) + 3y1G0(y1) 
  forsome = \G0y1.3y1 + G0(0) + 3y1G0(y1) + 3G0(y1) 
  nil = 3 
  or = \y0y1.y1 + 2y0 
  true = 0 

Using this interpretation, the requirements translate to:

  [[and(true, true)]] = 1 > 0 = [[true]] 
  [[and(_x0, false)]] = 1 + x0 > 0 = [[false]] 
  [[and(false, _x0)]] = 1 + x0 > 0 = [[false]] 
  [[or(true, _x0)]] = x0 >= 0 = [[true]] 
  [[or(_x0, true)]] = 2x0 >= 0 = [[true]] 
  [[or(false, false)]] = 0 >= 0 = [[false]] 
  [[forall(_F0, nil)]] = 9 + 11F0(3) > 0 = [[true]] 
  [[forall(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 11F0(3 + 3x1 + 3x2) > 1 + x1 + 3x2 + F0(x1) + 2F0(x2) + 3x2F0(x2) = [[and(_F0 _x1, forall(_F0, _x2))]] 
  [[forsome(_F0, nil)]] = 9 + F0(0) + 12F0(3) > 0 = [[false]] 
  [[forsome(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + F0(0) + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 12F0(3 + 3x1 + 3x2) > 2x1 + 3x2 + F0(0) + 2F0(x1) + 3x2F0(x2) + 3F0(x2) = [[or(_F0 _x1, forsome(_F0, _x2))]] 

We can thus remove the following rules:

  and(true, true) => true 
  and(X, false) => false 
  and(false, X) => false 
  forall(F, nil) => true 
  forall(F, cons(X, Y)) => and(F X, forall(F, Y)) 
  forsome(F, nil) => false 
  forsome(F, cons(X, Y)) => or(F X, forsome(F, 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(true, X) >? true 
  or(X, true) >? true 
  or(false, false) >? false 

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

The following interpretation satisfies the requirements:

  false = 0 
  or = \y0y1.3 + 3y0 + 3y1 
  true = 0 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  or(true, X) => true 
  or(X, true) => true 
  or(false, 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.