Proving independence with large cardinals? Suppose I want to prove some statement S is independent of ZFC.
Now instead of the usual approach of making models, I do the following:
- Take two large cardinal axioms L1 and L2
- Prove that ZFC + L1 $\implies$ S is true
- Prove that ZFC + L2 $\implies$ S is false
Then I argue that "Since ZFC + L1 implies S is true, then this means ZFC cannot prove S is false. Similarly, I have shown that ZFC cannot prove that S is true. Hence, S is independent of ZFC."
My questions are:
1) Does this approach work ? If so, has it actually been used ?
2) If not, what is the problem ? Is it the case that proving "S is consistent with ZFC" different from proving that "ZFC cannot prove S is false" ?
3) Or is the worry that I am adding axioms which might themselves be inconsistent with ZFC ? I realize that L1 and L2 could have been any two axioms, but I deliberately mentioned large cardinal axioms as they are generally believed to be consistent with ZFC. 
 A: The fundamental flaw with this approach is this part:

$\text{-}$ Take two large cardinal axioms L1 and L2

There is no singular formal definition of large cardinals, but in general, the fundamental property of large cardinals is:


*

*They form a linear hierarchy.

*Any two large cardinals are compatible with each other, assuming a sufficiently large large cardinal.
Here is the intuition behind 2. Given any large cardinal $\text{L}1$, $\text{L}1(\kappa)$ fundamentally asserts that $\kappa$ is so big that "[insert axiom]." It seems strange that there could be a number so big that "[insert axiom 1]," and there can be a number so big that "[insert axiom 2]," but not both at the same time.
On the other, hand there is a technique that goes like this: Take a large cardinal $\text{L}1$ and an axiom $S$. Then:


*$\text{L1}\rightarrow\text{Con}(\text{ZFC}+S)$

*$\text{L}1\rightarrow\neg S$
$S$ is the axiom $\exists U(V=L[U]\land U\text{ is a measure over a measurable cardinal})$. Let $\text{L}1$ be the assertion that there are two measurable cardinals. Then, by taking a witness to $0^\dagger$ exists, we can get a model $L[U]=V^{L[U]}$. But, because there are two measurable cardinals, $V\neq L[U]$.
