Independence of answers to queries sent to a random oracle Assume we have an algorithm which asks random oracle $\mathcal{O}$ $Q$ queries $u_1, \ldots, u_Q$. All queries are unique, $u_i \neq u_j$ for $i \neq j$. Queries $u_i$ are random variables, too.

What is the easiest (mathematically correct) way to see that the random variable $(\mathcal{O}(u_1),\ldots,\mathcal{O}(u_Q))$ has uniform distribution -- or -- that $\mathcal{O}(u_i)$ are independent random variables?

It looks like that the standard theorems covered in textbooks - like this one in Stinson - cannot be directly applied.

EDIT:

Here is more formal and simplified version of my question: $A$ is set of all random tapes (or private keys or something), $B$ is is set of all functions from, lets say, $\mathbb{Z}_q$ to $\mathbb{Z}_q$. $\mathcal{A}$ is some algorithm that has parameters $\omega \in A$ and random oracle $\mathcal{O}$ from $B$. We pick elements from $A$ and $B$ with uniform distribution.

For example, assume I want to calculate the probability
$
  P((\omega, \mathcal{O}) \in A\times B| \mathcal{A}(\omega, \mathcal{O}) \text{ succeeds})
$

This can be rewritten as

$
  P((\omega, \mathcal{O}) \in A\times B| \mathcal{A}(\omega, \mathcal{O}(u_1), \ldots, \mathcal{O}(u_Q)) \text{ succeeds})
$

And then I have a Stinson's theorem which can be replaced by some other  other "textbook" or citeable theorem.

The main goal is to prove $(\omega, \mathcal{O}(u_1), \ldots, \mathcal{O}(u_Q))$ has uniform distribution by citing something.
Firstly, note that saying that the queries $u_1,\ldots,u_Q$ are distinct implies that they are not independent, for example, if $u_i=a$ then $u_j\neq a$ for all $j&gt;i$.

Secondly, your argument indeed follows from the definition of a Random Oracle found in Stinson's book. Let $\mathcal{O}:\mathcal{X}\rightarrow \mathcal{Y}$ be a random oracle. That means that for every query $u$ that has not been queried before, it holds that $Pr[\mathcal{O}(u)=y]=1/|\mathcal{Y}|$ for every $y\in \mathcal{Y}$. Now, since $u_1,\ldots,u_Q$ are all distinct it follows that 

For all $y\in\mathcal{Y}$ and $i\in[Q]:  Pr[\mathcal{O}(u_i)=y]=\frac{1}{|\mathcal{Y}|}$ 

$\Rightarrow$ For all $(y_1,\ldots,y_Q)\in\mathcal{Y^Q}$:  $Pr[\mathcal{O}(u_1)=y_1,\ldots, \mathcal{O}(u_Q)=y_Q]= \prod_{i=1}^Q Pr[\mathcal{O}(u_i)=y_i]=\frac{1}{|\mathcal{Y}|^Q} $