
[;1m  substitution(SetFun, Set1)[0m

  Returns a function, the domain of which is [;;4mSet1[0m. The value of an
  element of the domain is the result of applying [;;4mSetFun[0m to the
  element.

    1> L = [{a,1},{b,2}].
    [{a,1},{b,2}]
    2> sofs:to_external(sofs:projection(1,sofs:relation(L))).
    [a,b]
    3> sofs:to_external(sofs:substitution(1,sofs:relation(L))).
    [{{a,1},a},{{b,2},b}]
    4> SetFun = {external, fun({A,_}=E) -> {E,A} end},
    sofs:to_external(sofs:projection(SetFun,sofs:relation(L))).
    [{{a,1},a},{{b,2},b}]

  The relation of equality between the elements of {a,b,c}:

    1> I = sofs:substitution(fun(A) -> A end, sofs:set([a,b,c])),
    sofs:to_external(I).
    [{a,a},{b,b},{c,c}]

  Let [;;4mSetOfSets[0m be a set of sets and [;;4mBinRel[0m a binary relation.
  The function that maps each element [;;4mSet[0m of [;;4mSetOfSets[0m onto the
  image of [;;4mSet[0m under [;;4mBinRel[0m is returned by the following
  function:

    images(SetOfSets, BinRel) ->
       Fun = fun(Set) -> sofs:image(BinRel, Set) end,
       sofs:substitution(Fun, SetOfSets).

  External unordered sets are represented as sorted lists. So,
  creating the image of a set under a relation R can traverse all
  elements of R (to that comes the sorting of results, the image).
  In [;;4mimage/2[0m, [;;4mBinRel[0m is traversed once for each element of [;;4m[0m
  [;;4mSetOfSets[0m, which can take too long. The following efficient
  function can be used instead under the assumption that the image
  of each element of [;;4mSetOfSets[0m under [;;4mBinRel[0m is non-empty:

    images2(SetOfSets, BinRel) ->
       CR = sofs:canonical_relation(SetOfSets),
       R = sofs:relative_product1(CR, BinRel),
       sofs:relation_to_family(R).
