TITLE: Why relativization can't solve NP !=P?
QUESTION [21 upvotes]: If this problem is really stupid, please close it. But I really wanna get some answer for it. And I learnt computational complexity by reading books only.
When I learnt to the topic of relativization and oracle machines, I read the following theorem:

There exist oracles A, B such that $P^A = NP^A$ and $P^B \neq NP^B$.

And then the book said because of this, we can't solve the problem of NP = P by using relativization. But I think what it implies is that $NP \neq P$. The reasoning is like this:
First of all, it is quite easy to see that:
$$A = B \Leftrightarrow \forall \text{oracle O, }A^O = B^O$$
Though I think it is obvious, I still give a proof to it:
A simple proof of NP != P ?
And the negation of it is:
$$A \neq B \Leftrightarrow \exists \text{oracle O such that } A^O \neq B^O$$
Therefore since there is an oracle B such that:
$$ NP^B \neq P^B$$
we can conclude that $ NP \neq P $
What's the problem with the above reasoning?

REPLY [4 votes]: And of course, this is exactly what happens in the $\mathrm{IP}^A \ne \mathrm{PSPACE}^A$ proof -- IP is defined by its access to outside information, which can be considered akin to an oracle -- while PSPACE does not.  In the most cursory sense, both the IP-verifier and the IP-prover have access to an oracle of random bits.  PSPACE-bound machines in general do not.  A cursory way of looking at this, without even reference to the proof (which frankly I forget) is that granting IP an oracle of random bits would by definition do noting to it.  But granting PSPACE an oracle of random bits could improve its computational ability.