Different versions of Einstein's equivalence principle As I understand it, there are two versions of Einstein's equivalence principle. The first states that 
"Locally, a frame in free-fall in a gravitational field is equivalent to an inertial frame in space in the absence of a gravitational field".
The second states that 
"Locally, a uniformly accelerating frame in space (in the absence of a gravitational field) is equivalent to a frame at rest in a uniform gravitational field"
I realise that these two statements should be equivalent (no pun intended), but it doesn't seem immediately obvious to me and I'm hoping someone can explain?! (Is it simply that a frame in a gravitational field, accelerating purely due to the influence of gravity is equivalent to an inertial frame in free space, in the absence of gravity. This statement can be reversed, that is to say, a frame at rest in a gravitational field, purely due to the influence of gravity, is equivalent to a uniformly accelerating frame in free space?)
Furthermore, in the first case, I have seen elementary arguments where one considers a lift in free-fall, as follows: According to the observer on the ground, the net force acting on a particle within the lift is given by $m\mathbf{g}+\mathbf{F}=m\mathbf{a}$ (where $\mathbf{F}$ is the net force acting on the particle, other than gravity). The acceleration, $\mathbf{a}'$ measured by an observer in the lift is related to the acceleration, $\mathbf{a}$ measured by the observer on the ground by $\mathbf{a}=\mathbf{a}_{0}+\mathbf{a}'$, where $\mathbf{a}_{0}$ is the acceleration of the lift, and since it is in free-fall, $\mathbf{a}_{0}=\mathbf{g}$. Hence, $$m\mathbf{g}+\mathbf{F}=m\mathbf{g}+m\mathbf{a}'\quad\Rightarrow\quad\mathbf{F}=m\mathbf{a}'$$ and so the observer in the lift measures the same force acting on the particle as the observer on the ground, hence they are also in an inertial frame.
The problem I find with this argument is that the observer on the ground is not in an inertial frame themselves (since Earth is not an inertial frame). I'm I missing the point here though? Is it simply that this argument is used in the framework of Newtonian mechanics, where the Earth is considered an inertial frame, and so one can use this argument to motivate the equivalence principle (a believe this kind of argument was used by Einstein initially to convince himself of the equivalence principle)?!
 A: The equivalence principle in modern terminology comes in 2, and some say 3, forms: weak, Einstenian, and strong. See them in the Wikipedia site at:
https://en.m.wikipedia.org/wiki/Equivalence_principle
Your statements are a little confusing, but your question that the earth is not an inertial frame, although right, for those experiments stated by you) is considered (and to an approximation is) an inertial frame. And if you assume that g for an elevator is only due to earth, you are basically assuming that. I.e., forget about the earth accelerating. 
Read the wiki paper, easier to follow, and does a little history but discusses the 2 (or 3) modern versions. Yours is more or less the weak equivalent principle (where the earth is not mentioned at all, not the issue). Maybe yours includes the Einstenian which is the weak plus a bit more. 
The basic idea is the universality of free fall. But there aremany equivalent ways of stating it, see them in the wiki. Also leads to the equivalence of inertial and gravitational mass 
A: The two are equivalent, but your first version is the simplest.  The idea is to start from Newtonian gravity, and consider the fact that, at a given point in space, the gravitational acceleration is the same no matter what object is placed there.  Now expand that point to a small region of space, so that multiple objects can be in it.  Let them be free-falling.  In the limit as the region is small enough, all those objects still have the same acceleration, hence their relative velocities are not changing.  But that means, if you're an observer in this region, you can't tell that there is gravity; your experience is indistinguishable from inertial motion.
To get the second version you cite, you have to add an additional force.  To the inertial observer, you have to add the force of the rocket accelerating them and all objects around them.  To the observer in the gravitational field, you have to add the force of the ground holding them up.  Thus "inertial" gets replaced with "accelerating", and "free-falling" becomes "at rest".
Einstein's breakthrough, aided by Grossmann, was to realize that, if free-fall and inertial motion are already equivalent in small enough (infinitesimal) regions of Newtonian/Galilean spacetime, then maybe the way to reconcile gravity with special relativity was to suppose that they in fact followed the same mathematical law.  But since inertial paths are straight, the only way deviant free-fall paths could be "straight" is if they were geodesics of a curved spacetime.
