/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 : [] --> c 
    cons : [c * b] --> b 
    map : [c -> c * b] --> b 
    nil : [] --> b 
    node : [a * b] --> c 
    plus : [c * c] --> c 
    s : [c] --> c 
    size : [] --> c -> c 
    sum : [b] --> c 

  Rules:

    map(f, nil) => nil 
    map(f, cons(x, y)) => cons(f x, map(f, y)) 
    sum(cons(x, y)) => plus(x, sum(y)) 
    size node(x, y) => s(sum(map(size, y))) 
    plus(0, x) => 0 
    plus(s(x), y) => s(plus(x, 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]):

  map(F, nil) >? nil 
  map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 
  sum(cons(X, Y)) >? plus(X, sum(Y)) 
  size node(X, Y) >? s(sum(map(size, Y))) 
  plus(0, X) >? 0 
  plus(s(X), Y) >? s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

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

Using this interpretation, the requirements translate to:

  [[map(_F0, nil)]] = 6 + F0(0) + 3F0(3) > 3 = [[nil]] 
  [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 4x2 + F0(0) + 2x2F0(3 + x1 + 2x2) + 3F0(3 + x1 + 2x2) + x1F0(3 + x1 + 2x2) > 3 + x1 + 4x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[cons(_F0 _x1, map(_F0, _x2))]] 
  [[sum(cons(_x0, _x1))]] = 3 + x0 + 2x1 > x0 + x1 = [[plus(_x0, sum(_x1))]] 
  [[size node(_x0, _x1)]] = 4 + x0 + 3x1 > 1 + 3x1 = [[s(sum(map(size, _x1)))]] 
  [[plus(0, _x0)]] = x0 >= 0 = [[0]] 
  [[plus(s(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[s(plus(_x0, _x1))]] 

We can thus remove the following rules:

  map(F, nil) => nil 
  map(F, cons(X, Y)) => cons(F X, map(F, Y)) 
  sum(cons(X, Y)) => plus(X, sum(Y)) 
  size node(X, Y) => s(sum(map(size, 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]):

  plus(0, X) >? 0 
  plus(s(X), Y) >? s(plus(X, Y)) 

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

The following interpretation satisfies the requirements:

  0 = 0 
  plus = \y0y1.3 + y1 + 3y0 
  s = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[plus(0, _x0)]] = 3 + x0 > 0 = [[0]] 
  [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] 

We can thus remove the following rules:

  plus(0, X) => 0 
  plus(s(X), Y) => s(plus(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.