/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:

    app : [list * list] --> list 
    cons : [nat * list] --> list 
    map : [nat -> nat * list] --> list 
    nil : [] --> list 
    reverse : [list] --> list 
    shuffle : [list] --> list 

  Rules:

    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    reverse(nil) => nil 
    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
    shuffle(nil) => nil 
    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(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]):

  app(nil, X) >? X 
  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  reverse(nil) >? nil 
  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
  shuffle(nil) >? nil 
  shuffle(cons(X, Y)) >? cons(X, shuffle(reverse(Y))) 
  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.2 + y0 + y1 
  map = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
  nil = 0 
  reverse = \y0.y0 
  shuffle = \y0.2 + y0 

Using this interpretation, the requirements translate to:

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

We can thus remove the following rules:

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

  app(nil, X) >? X 
  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  reverse(nil) >? nil 
  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 
  shuffle(cons(X, Y)) >? cons(X, shuffle(reverse(Y))) 
  map(F, nil) >? nil 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 + 2y0 
  map = \G0y1.3 + 3y1 + G0(0) 
  nil = 0 
  reverse = \y0.y0 
  shuffle = \y0.2y0 

Using this interpretation, the requirements translate to:

  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[shuffle(cons(_x0, _x1))]] = 2 + 2x1 + 4x0 > 1 + 2x0 + 2x1 = [[cons(_x0, shuffle(reverse(_x1)))]] 
  [[map(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] 

We can thus remove the following rules:

  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
  map(F, nil) => nil 

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

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

  app(nil, X) >? X 
  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 
  reverse(nil) >? nil 
  reverse(cons(X, Y)) >? app(reverse(Y), cons(X, nil)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.2 + y0 + y1 
  cons = \y0y1.3 + y1 + 3y0 
  nil = 1 
  reverse = \y0.1 + 3y0 

Using this interpretation, the requirements translate to:

  [[app(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 5 + x1 + x2 + 3x0 >= 5 + x1 + x2 + 3x0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 4 > 1 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 10 + 3x1 + 9x0 > 7 + 3x0 + 3x1 = [[app(reverse(_x1), cons(_x0, nil))]] 

We can thus remove the following rules:

  app(nil, X) => X 
  reverse(nil) => nil 
  reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 

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

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

  app(cons(X, Y), Z) >? cons(X, app(Y, Z)) 

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

The following interpretation satisfies the requirements:

  app = \y0y1.y1 + 3y0 
  cons = \y0y1.1 + y0 + y1 

Using this interpretation, the requirements translate to:

  [[app(cons(_x0, _x1), _x2)]] = 3 + x2 + 3x0 + 3x1 > 1 + x0 + x2 + 3x1 = [[cons(_x0, app(_x1, _x2))]] 

We can thus remove the following rules:

  app(cons(X, Y), Z) => cons(X, app(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.