Set partition problem in Essence'

$Problem formulation from$   http://www.koalog.com/resources/samples/PartitionProblem.java.html
$"""$  This is a partition problem.
$Given the set S = {1, 2, ..., n},$  it consists in finding two sets A and B such that:
$<ul>$  <li>A U B = S,</li>
$<li>|A| = |B|,</li>$  <li>sum(A) = sum(B),</li>
$<li>sum_squares(A) = sum_squares(B).</li>$  </ul>
$"""$
$This Essence' model was created by Hakan Kjellerstrand, hakank@gmail.com .$ See also my Tailor/Essence' page: http://www.hakank.org/savile_row/ .
 Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/

language ESSENCE' 1.0

$For num_sets n must be a multiple of 4 for this to work.$ And - of course - num_sets must be a multiple of n.

letting n be 16
letting num_sets be 2

$one solution$ 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16
$1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2;$ ->
$1: {1,2,7,8,11,12,13,14},$ 2: {3,4,5,9,10,15,16}

$Essence' don't support sets so we represent the sets$ simply as a value in the a array.
find a: matrix indexed by [int(1..n)] of int(1..num_sets)

find sums: matrix indexed by [int(1..num_sets)] of int(0..n*n)
find sums_squared: matrix indexed by [int(1..num_sets)] of int(0..n*n*n*n)

such that
forall i: int(1..num_sets) .
sums[i] = (sum j: int(1..n) . j*(a[j]=i)) /\
sums_squared[i] = sum j: int(1..n) . j**2*(a[j]=i)
,
$same cardinality and sums forall i: int(2..num_sets) . (sum j: int(1..n) . a[j]=i-1) = (sum j: int(1..n) . a[j]=i) /\ sums[i-1] = sums[i] /\ sums_squared[i-1] = sums_squared[i] ,$ summetry breaking
a = 1