code: def foldl1(f: Callable[[T, T], T], xs: Iterable[T]) -> T:
      
  
      return reduce(f, xs)
docs: Returns the accumulated result of a binary function applied to elements
      of an iterable.
  
      .. math::
          foldl1(f, [x_0, x_1, x_2, x_3]) = f(f(f(f(x_0, x_1), x_2), x_3)
  
  
      Examples
      --------
      >>> from delphi.utils.fp import foldl1
      >>> foldl1(lambda x, y: x + y, range(5))
      10