/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:

    append : [a * a] --> a 
    concat : [a] --> a 
    cons : [a * a] --> a 
    flatten : [] --> a -> a 
    map : [a -> a * a] --> a 
    nil : [] --> a 
    node : [a * a] --> a 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    flatten node(x, y) => cons(x, concat(map(flatten, y))) 
    concat(nil) => nil 
    concat(cons(x, y)) => append(x, concat(y)) 
    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 

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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 
  concat(cons(X, Y)) >? append(X, concat(Y)) 
  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  concat = \y0.y0 
  cons = \y0y1.1 + y0 + y1 
  flatten = \y0.0 
  map = \G0y1.2 + y1 + G0(y1) + 2y1G0(y1) 
  nil = 0 
  node = \y0y1.3 + 3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) >= 3 + x1 + x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 
  [[concat(nil)]] = 0 >= 0 = [[nil]] 
  [[concat(cons(_x0, _x1))]] = 1 + x0 + x1 > x0 + x1 = [[append(_x0, concat(_x1))]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 

We can thus remove the following rules:

  map(F, nil) => nil 
  concat(cons(X, Y)) => append(X, concat(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, cons(X, Y)) >? cons(F X, map(F, Y)) 
  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 
  append(nil, X) >? X 
  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 

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

The following interpretation satisfies the requirements:

  append = \y0y1.3 + y1 + 3y0 
  concat = \y0.y0 
  cons = \y0y1.3 + y0 + y1 
  flatten = \y0.0 
  map = \G0y1.2y1 + G0(0) + 3y1G0(y1) 
  nil = 0 
  node = \y0y1.3 + 3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) > 3 + x1 + 2x2 + F0(0) + F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + 2x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 
  [[concat(nil)]] = 0 >= 0 = [[nil]] 
  [[append(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 

We can thus remove the following rules:

  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  append(nil, X) => X 
  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 

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

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

  flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 
  concat(nil) >? nil 

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

The following interpretation satisfies the requirements:

  concat = \y0.2y0 
  cons = \y0y1.y0 + y1 
  flatten = \y0.3 + 3y0 
  map = \G0y1.y1 + G0(0) 
  nil = 2 
  node = \y0y1.3 + 3y0 + 3y1 

Using this interpretation, the requirements translate to:

  [[flatten node(_x0, _x1)]] = 15 + 12x0 + 12x1 > 6 + x0 + 2x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 
  [[concat(nil)]] = 4 > 2 = [[nil]] 

We can thus remove the following rules:

  flatten node(X, Y) => cons(X, concat(map(flatten, Y))) 
  concat(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.