Is $(NP^{NP})^{NP} = NP^{(NP^{NP})}$?

Question

In the "last paragraph" of the "first page" of the following paper:

Vikraman Arvind, Johannes Köbler, Uwe Schöning, Rainer Schuler, "If NP Has Polynomial-Size Circuits, then MA = AM," Theoretical Computer Science, 1995.

I encountered a somewhat counter-intuitive claim:

$(\Sigma^P_2 \cap \Pi^P_2)^{NP} = \Sigma^P_3 \cap \Pi^P_3$

I think the identity above is deduced from the following:

$(\Sigma^P_2)^{NP} = \Sigma^P_3$

and

$(\Pi^P_2)^{NP} = \Pi^P_3$

The former is more simply written as $(NP^{NP})^{NP} = NP^{NP^{NP}}$, which is quite odd!

Edit: In light of Kristoffer's comment below, I'd like to add the following inspiring remark from Goldreich's complexity book (pp. 118-119):

It should be clear that the class $C_1^{C_2}$ can be defined for two complexity classes $C_1$ and $C_2$, provided that $C_1$ is associated with a class of standard machines that generalizes naturally to a class of oracle machines. Actually, the class $C_1^{C_2}$ is not defined based on the class $C_1$ but rather by analogy to it. Specifically, suppose that $C_1$ is the class of sets that are recognizable (or rather accepted) by machines of a certain type (e.g., deterministic or non-deterministic) with certain resource bounds (e.g., time and/or space bounds). Then, we consider analogous oracle machines (i.e., of the same type and with the same resource bounds), and say that $S \in C_1^{C_2}$ if there exists an adequate oracle machine $M_1$ (i.e., of this type and resource bounds) and a set $S_2 \in C_2$ such that $M_1^{S_2}$ accepts the set $S$.

Answer 1

${\Sigma_2^P}^{NP}$ is the set of language decided by an alternating turing machine in existential, and then universal state, with an oracle in NP. Both the universal and the existantial part can querye NP.

Hence, in this case you decided to write this as $(NP^{NP})^{A}$ then the way you should think of it is as $(NP^{NP^A\cup A})$ (by $\cup$ I mean an oracle either to $A$ or to an $NP^A$ language).

Hence ${\Sigma_2^P}^{NP}$ is equal to $(NP^{(NP^{NP})})^{NP}$ which is certainly equal to $(NP^{NP^{NP}})$ since every query you could make to the $NP$ oracle, you could make it to the $NP^{NP}$ oracle.

Answer 2

From Arora and Barak (p. 102) theorem 5.12: "For every $i\geq 2$, $\sum_i^p=NP^{\sum_{i-1}SAT}$". Remember that $\sum_{i}SAT$ is the QBF formula with $i$ alternations which is complete for $\sum_i^p$. Then $\sum_2^p=NP^{SAT}$ and given that SAT is NP-complete you just write $\sum_2^p=NP^{NP}$, so far so good. Extending this notation to $i=3$ you get $NP^{NP^{NP}}$, but the last two "NPs" are just an oracle for the language $\sum_{2}SAT$ with at most 2 alternations. It seems to me that its just a shorthand notation for oracle access.