/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 : [c * d] --> d 
    f : [a] --> a 
    false : [] --> b 
    filter : [c -> b * d] --> d 
    filter2 : [b * c -> b * c * d] --> d 
    g : [a] --> a 
    map : [c -> c * d] --> d 
    nil : [] --> d 
    true : [] --> b 

  Rules:

    f(f(x)) => g(f(x)) 
    g(g(x)) => f(x) 
    map(h, nil) => nil 
    map(h, cons(x, y)) => cons(h x, map(h, y)) 
    filter(h, nil) => nil 
    filter(h, cons(x, y)) => filter2(h x, h, x, y) 
    filter2(true, h, x, y) => cons(x, filter(h, y)) 
    filter2(false, h, x, y) => filter(h, 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]):

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(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:

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

Using this interpretation, the requirements translate to:

  [[f(f(_x0))]] = 4x0 >= 4x0 = [[g(f(_x0))]] 
  [[g(g(_x0))]] = 4x0 >= 2x0 = [[f(_x0)]] 
  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 5 + 3x1 + 3x2 + 2F0(1 + x1 + x2) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + F0(x2) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[filter(_F0, nil)]] = F0(0) >= 0 = [[nil]] 
  [[filter(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(1 + x1 + x2) > 3x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 
  [[filter2(true, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 1 + x1 + 3x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 
  [[filter2(false, _F0, _x1, _x2)]] = 6 + x1 + 3x2 + F0(0) + 2x2F0(x2) > 3x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] 

We can thus remove the following rules:

  map(F, nil) => nil 
  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]):

  f(f(X)) >? g(f(X)) 
  g(g(X)) >? f(X) 
  filter(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  f = \y0.2 + 3y0 
  filter = \G0y1.3 + 3y1 + G0(0) 
  g = \y0.2 + 2y0 
  nil = 0 

Using this interpretation, the requirements translate to:

  [[f(f(_x0))]] = 8 + 9x0 > 6 + 6x0 = [[g(f(_x0))]] 
  [[g(g(_x0))]] = 6 + 4x0 > 2 + 3x0 = [[f(_x0)]] 
  [[filter(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  f(f(X)) => g(f(X)) 
  g(g(X)) => f(X) 
  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.