/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:

    0 : [] --> b 
    cons : [b * a] --> a 
    inc : [] --> a -> a 
    map : [b -> b] --> a -> a 
    nil : [] --> a 
    plus : [b] --> b -> b 
    s : [b] --> b 

  Rules:

    plus(0) x => x 
    plus(s(x)) y => s(plus(x) y) 
    map(f) nil => nil 
    map(f) cons(x, y) => cons(f x, map(f) y) 
    inc => map(plus(s(0))) 

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]):

  plus(0) X >? X 
  plus(s(X)) Y >? s(plus(X) Y) 
  map(F) nil >? nil 
  map(F) cons(X, Y) >? cons(F X, map(F) Y) 
  inc >? map(plus(s(0))) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[plus(0) _x0]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0)) _x1]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[s(plus(_x0) _x1)]] 
  [[map(_F0) nil]] = 3 + F0(0) + 3F0(3) >= 3 = [[nil]] 
  [[map(_F0) cons(_x1, _x2)]] = 3 + x1 + x2 + F0(0) + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 3 + x1 + x2 + F0(0) + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0) _x2)]] 
  [[inc]] = \y0.3 + 3y0 > \y0.2 + 2y0 = [[map(plus(s(0)))]] 

We can thus remove the following rules:

  plus(s(X)) Y => s(plus(X) Y) 
  inc => map(plus(s(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]):

  plus(0, X) >? X 
  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:

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

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 6 + x0 > x0 = [[_x0]] 
  [[map(_F0, nil)]] = 3 + 4F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 12 + 3x1 + 3x2 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) > 6 + x1 + 3x2 + F0(x1) + 2F0(0) + 2F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 

We can thus remove the following rules:

  plus(0, X) => X 
  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 

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.