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


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YES
We consider the system theBenchmark.

  Alphabet:

    append : [b * b] --> b 
    cons : [a * b] --> b 
    map : [a -> a * b] --> b 
    nil : [] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    append(append(x, y), z) => append(x, append(y, z)) 
    map(f, append(x, y)) => append(map(f, x), map(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]):

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

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[append(nil, _x0)]] = 2 + x0 > x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 4 + x0 + x1 + x2 >= 4 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[map(_F0, nil)]] = 1 + 2F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 7 + 3x1 + 3x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 6F0(2 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2x2F0(x2) + 2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = 4 + x0 + x1 + x2 >= 4 + x0 + x1 + x2 = [[append(_x0, append(_x1, _x2))]] 
  [[map(_F0, append(_x1, _x2))]] = 7 + 3x1 + 3x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 6F0(2 + x1 + x2) > 4 + 3x1 + 3x2 + 2x1F0(x1) + 2x2F0(x2) + 2F0(x1) + 2F0(x2) = [[append(map(_F0, _x1), map(_F0, _x2))]] 

We can thus remove the following rules:

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

  append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 
  append(append(X, Y), Z) >? append(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 
  cons = \y0y1.2 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[append(cons(_x0, _x1), _x2)]] = 9 + x2 + 3x0 + 3x1 > 5 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 
  [[append(append(_x0, _x1), _x2)]] = 12 + x2 + 3x1 + 9x0 > 6 + x2 + 3x0 + 3x1 = [[append(_x0, append(_x1, _x2))]] 

We can thus remove the following rules:

  append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
  append(append(X, Y), Z) => append(X, append(Y, Z)) 

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.