/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:

    0 : [] --> b 
    cons : [b * a] --> a 
    curry : [b -> b -> b * b] --> b -> b 
    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) 
    curry(f, x) y => f x y 
    inc => map(curry(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) 
  curry(F, X) Y >? F X Y 
  inc >? map(curry(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 
  curry = \G0y1y2.y1 + 2G0(y1,y2) 
  inc = \y0.3 + 3y0 
  map = \G0y1.1 + G0(y1) + 2y1G0(y1) 
  nil = 3 
  plus = \y0y1.0 
  s = \y0.y0 

Using this interpretation, the requirements translate to:

  [[plus 0 _x0]] = x0 >= x0 = [[_x0]] 
  [[plus s(_x0) _x1]] = x0 + x1 >= x0 + x1 = [[s(plus _x0 _x1)]] 
  [[map(_F0) nil]] = 4 + 7F0(3) > 3 = [[nil]] 
  [[map(_F0) cons(_x1, _x2)]] = 4 + x1 + x2 + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 7F0(3 + x1 + x2) >= 4 + x1 + x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0) _x2)]] 
  [[curry(_F0, _x1) _x2]] = x1 + x2 + 2F0(x1,x2) >= x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] 
  [[inc]] = \y0.3 + 3y0 > \y0.1 = [[map(curry(plus, s(0)))]] 

We can thus remove the following rules:

  map(F) nil => nil 
  inc => map(curry(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 
  plus(s(X), Y) >? s(plus(X, Y)) 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  curry(F, X, Y) >? F X 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 
  curry = \G0y1y2.3 + y1 + y2 + G0(y1,y2) 
  map = \G0y1.3y1 + G0(0) + 2y1G0(y1) 
  plus = \y0y1.3 + y1 + 3y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 12 + x0 > x0 = [[_x0]] 
  [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 
  [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + F0(0) + 2x1F0(3 + x1 + x2) + 2x2F0(3 + x1 + x2) + 6F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[curry(_F0, _x1, _x2)]] = 3 + x1 + x2 + F0(x1,x2) > x1 + x2 + F0(x1,x2) = [[_F0 _x1 _x2]] 

We can thus remove the following rules:

  plus(0, X) => X 
  plus(s(X), Y) => s(plus(X, Y)) 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  curry(F, X, Y) => F X 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.