If I try to use a with statement inside of another with statement using a value from the outer with statement, the compiler uses the inner value to retrieve the value to use as the inner value. For example,

with something!
  with .other_thing!
    \do_the_thing!

results in the following code

do
  local _with_0 = something()
  do
    local _with_1 = _with_1.other_thing()
    _with_1:do_the_thing()
  end
  return _with_0
end

I compiled this code using the latest version of the compiler through the programmatic API.