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


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YES
We consider the system theBenchmark.

  Alphabet:

    cons : [a * c] --> c 
    dropWhile : [a -> b * c] --> c 
    if : [b * c * c] --> c 
    nil : [] --> c 
    takeWhile : [a -> b * c] --> c 
    true : [] --> b 

  Rules:

    if(true, x, y) => x 
    if(true, x, y) => y 
    takeWhile(f, nil) => nil 
    takeWhile(f, cons(x, y)) => if(f x, cons(x, takeWhile(f, y)), nil) 
    dropWhile(f, nil) => nil 
    dropWhile(f, cons(x, y)) => if(f x, dropWhile(f, y), cons(x, 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]):

  if(true, X, Y) >? X 
  if(true, X, Y) >? Y 
  takeWhile(F, nil) >? nil 
  takeWhile(F, cons(X, Y)) >? if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, nil) >? nil 
  dropWhile(F, cons(X, Y)) >? if(F X, dropWhile(F, Y), cons(X, Y)) 

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

The following interpretation satisfies the requirements:

  cons = \y0y1.2 + y0 + 2y1 
  dropWhile = \G0y1.3 + 2y1 + 2y1G0(y1) + 2G0(0) 
  if = \y0y1y2.y0 + y1 + y2 
  nil = 0 
  takeWhile = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
  true = 3 

Using this interpretation, the requirements translate to:

  [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x0 = [[_x0]] 
  [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x1 = [[_x1]] 
  [[takeWhile(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[takeWhile(_F0, cons(_x1, _x2))]] = 6 + 3x1 + 6x2 + F0(0) + 2x1F0(2 + x1 + 2x2) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 2 + 2x1 + 6x2 + F0(x1) + 2F0(0) + 4x2F0(x2) = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] 
  [[dropWhile(_F0, nil)]] = 3 + 2F0(0) > 0 = [[nil]] 
  [[dropWhile(_F0, cons(_x1, _x2))]] = 7 + 2x1 + 4x2 + 2x1F0(2 + x1 + 2x2) + 2F0(0) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 5 + 2x1 + 4x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _x2))]] 

We can thus remove the following rules:

  if(true, X, Y) => X 
  if(true, X, Y) => Y 
  takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) 
  dropWhile(F, nil) => nil 
  dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(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]):

  takeWhile(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  nil = 0 
  takeWhile = \G0y1.3 + 3y1 + G0(0) 

Using this interpretation, the requirements translate to:

  [[takeWhile(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  takeWhile(F, nil) => nil 

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.