Echoing everything @Aetherus said, with a similar example but just using recursion:

defmodule Match do
  @a [1,2,3,4,5]
  @b [5,4,3,2,1]

  # Elixir has lists, not arrays, and therefore
  # its O(n) to access elements and typically this
  # is not what you want. For this example, reversing
  # the second list once is far more efficient
  def reverse(a \\ @a, b \\ @b) do
    a
    |> :lists.reverse()
    |> reverse(b, 0)
  end

  # length(a) == 0
  def reverse([], _, count) do
    count
  end

  # the head of each list is the same fulfilling
  # a[length(a)-1] == b[I]
  # notice we are pattern matching the head of each list
  # and proceeding only if they are the same
  def reverse([first | a_rest], [first | b_rest], count) do
    reverse(a_rest, b_rest, count + 1)
  end

  # When the head of the lists no longer match we return
  def reverse(_, _, count) do
    count
  end
end

TL; DR you can’t.

In Elixir, everything is immutable, so even if there is a.pop(), it won’t work as you expected.

According to your code, I guess what you want to do is find the index of the first pair of identical elements from reversed a and b. If I’m right, then the code can be

a = [1,2,3,4,5]
b = [5,4,3,2,1]

i =
  a
  |> Enum.reverse()
  |> Enum.zip(b)
  |> Enum.find_index(fn {ea, eb} -> ea == eb end)