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

  Rules:

    plus(0) x => x 
    plus(s(x)) y => s(plus(x) y) 
    inc(x) => map(plus(s(0)), x) 
    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]):

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

Using this interpretation, the requirements translate to:

  [[plus(0) _x0]] = x0 >= x0 = [[_x0]] 
  [[plus(s(_x0)) _x1]] = x1 + 2x0 >= x1 + 2x0 = [[s(plus(_x0) _x1)]] 
  [[inc(_x0)]] = 3 + 3x0 > 2 + x0 = [[map(plus(s(0)), _x0)]] 
  [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 5 + x2 + 2x1 + F0(0) + 2F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 

We can thus remove the following rules:

  inc(X) => map(plus(s(0)), X) 
  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]):

  plus(0, X) >? X 
  plus(s(X), Y) >? s(plus(X, Y)) 
  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 = \G0y1.3y1 + 2G0(0) + 2G0(y1) + 3y1G0(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 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) > 3 + 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 
  plus(s(X), Y) => s(plus(X, Y)) 
  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.