/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:

    !facdot : [a * a] --> a 
    1 : [] --> a 
    cons : [c * d] --> d 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    i : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    !facdot(1, x) => x 
    !facdot(x, 1) => x 
    !facdot(i(x), x) => 1 
    !facdot(x, i(x)) => 1 
    !facdot(i(x), !facdot(x, y)) => y 
    !facdot(x, !facdot(i(x), y)) => y 
    i(1) => 1 
    i(i(x)) => x 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    filter(f, nil) => nil 
    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
    filter2(true, f, x, y) => cons(x, filter(f, y)) 
    filter2(false, f, x, y) => filter(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]):

  !facdot(1, X) >? X 
  !facdot(X, 1) >? X 
  !facdot(i(X), X) >? 1 
  !facdot(X, i(X)) >? 1 
  !facdot(i(X), !facdot(X, Y)) >? Y 
  !facdot(X, !facdot(i(X), Y)) >? Y 
  i(1) >? 1 
  i(i(X)) >? X 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  filter(F, nil) >? nil 
  filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) >? filter(F, Y) 

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

The following interpretation satisfies the requirements:

  !facdot = \y0y1.3 + 3y0 + 3y1 
  1 = 0 
  cons = \y0y1.2 + y0 + y1 
  false = 3 
  filter = \G0y1.2y1 + G0(0) + 2y1G0(y1) 
  filter2 = \y0G1y2y3.y0 + y2 + 2y3 + G1(0) + 2y3G1(y3) 
  i = \y0.3 + 3y0 
  map = \G0y1.3y1 + G0(0) + G0(y1) + 2y1G0(y1) 
  nil = 0 
  true = 3 

Using this interpretation, the requirements translate to:

  [[!facdot(1, _x0)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[!facdot(_x0, 1)]] = 3 + 3x0 > x0 = [[_x0]] 
  [[!facdot(i(_x0), _x0)]] = 12 + 12x0 > 0 = [[1]] 
  [[!facdot(_x0, i(_x0))]] = 12 + 12x0 > 0 = [[1]] 
  [[!facdot(i(_x0), !facdot(_x0, _x1))]] = 21 + 9x1 + 18x0 > x1 = [[_x1]] 
  [[!facdot(_x0, !facdot(i(_x0), _x1))]] = 39 + 9x1 + 30x0 > x1 = [[_x1]] 
  [[i(1)]] = 3 > 0 = [[1]] 
  [[i(i(_x0))]] = 12 + 9x0 > x0 = [[_x0]] 
  [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 6 + 3x1 + 3x2 + F0(0) + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 5F0(2 + x1 + x2) > 2 + x1 + 3x2 + F0(0) + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 4 + 2x1 + 2x2 + F0(0) + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 4F0(2 + x1 + x2) > 2x1 + 2x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2 + x1 + 2x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 3 + x1 + 2x2 + F0(0) + 2x2F0(x2) > 2x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  !facdot(1, X) => X 
  !facdot(X, 1) => X 
  !facdot(i(X), X) => 1 
  !facdot(X, i(X)) => 1 
  !facdot(i(X), !facdot(X, Y)) => Y 
  !facdot(X, !facdot(i(X), Y)) => Y 
  i(1) => 1 
  i(i(X)) => X 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 
  filter2(true, F, X, Y) => cons(X, filter(F, Y)) 
  filter2(false, F, X, Y) => filter(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]):

  map(F, nil) >? nil 
  filter(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

  map(F, nil) => nil 
  filter(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.