FWIW, you’ll get much better responses if you describe what this code is supposed to do.

Tracing through an example by hand, it will return the whole input list if every given MapSet intersects with every other one. Non-intersecting sets will be combined.

Some general notes on structure:

• the second argument is always patterned-matched against either [] or [item]. Consider removing the useless list wrapping and use nil vs item instead

• +1 to using if over a cond with one non-default branch

• consider using MapSet.disjoint?/2 - there’s probably a tiny performance benefit to not constructing the whole result of MapSet.intersection, but more importantly it’s clearer what the intent of the check is

Why so complicate?

iex> m1 = MapSet.new([1, 2, 3])
#MapSet<[1, 2, 3]>
iex> m2 = MapSet.new([2, 3, 4])
#MapSet<[2, 3, 4]>
iex> MapSet.intersection m1, m2
#MapSet<[2, 3]>

… or maybe I didn’t get what You want

UPDATE: Well, I didn’t get what You want to do… better use data and expected result in that case.
Also, I would not write it like this… first the main entry function, then the rest private…

defmodule Koko do
  def meld(list) do
    ....
  end

  defp do_meld(a, b, c, d) do
    ....
  end
end

With your code, You need to start to read from the bottom.

You might use Enum.reduce/3 in the following way (untested, but it should work):

Enum.reduce(sets, [], fn set, sets ->
  case Enum.split_with(sets, &MapSet.disjoint?(&1, set)) do
    {_, []} ->  # not a single join, totally alien, appending
      [set | sets]
    {disjoined, joined} ->  # ⇓⇓⇓ here is a trick ⇓⇓⇓
      [Enum.reduce(joined, set, &MapSet.union/2) | disjoined]
  end
end)

The only interesting thing here is that once we discovered all the sets joined the tested one, all of them are to be joined.