$A$ is indefinite iff $A$ fits none of the above criteria. Equivalently, $A$ has both positive and negative eigenvalues. Also equivalently, $x^TAx$ is positive for at least one $x$ and negative for at least another $x$.
Note that the leading principal minors refer to the determinants of the northwest-corner submatrices, and are merely a subset of all the principal minors.
Now, suppose that a symmetric $n\times n$ matrix $M$ is neither positive definite nor negative definite. $$$$ From the facts highlighted above, and possibly using linear algebra, then is statement (2) true? If not, is at least statement (1) true? I have seen both assertions separately in different texts (e.g. http://people.ds.cam.ac.uk/iar1/teaching/Hessians-DefinitenessTutorial.pdf and http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/BlumeSimonCh16.PDF), but am unable to prove either:
(1) If $M$'s leading principal minors are all nonzero, then $M$ is indefinite.
(2) If $M$ has some nonzero leading principal minor, then $M$ is indefinite.
EDIT: (1) can actually be simplified:
(1) If det $M$ is nonzero, then $M$ is indefinite.
We don't need to check all the leading principal minors because once det M is nonzero, we can immediately deduce that M has no zero eigenvalues, and since it is also given that M is neither positive definite nor negative definite, then M can only be indefinite.