TITLE: Oracle queries asked in parallel
QUESTION [7 upvotes]: Definition: Assume that $\phi(q)$ is of the form $\exists y \leq 2^{p(n)} \varphi(q,y)$, where $p$ is a polynomial and $n = |q|$ (i.e. $n$ is the length of the binary representation of $q$). Then a witness query $q$ to $\phi$ returns a witness $y \leq 2^{p(n)}$ satisfying $\varphi(q,y)$, if such exists, and returns $2^{p(n)} + 1$, otherwise. 
Question: Let $f$ be a function in $FP^{\Sigma^p_1}[wit]$, i.e. $f$ is computed by a polynomial time Turing machine $M$ that makes polynomially many witness queries to a $\Sigma^p_1$ oracle $\phi(q) \equiv \exists y \leq 2^{p(n)} \varphi(q,y)$. 
My question is the following: Can we somehow construct a polynomial time Turing machine $M_1$ which makes polynomially many witness queries to $\phi(q)$
in parallel (i.e. all the queries to $\phi$ come first, then the sequence of replies  to the queries come after) and such that $M^{\phi}_1(x) = M^{\phi}(x)$.

REPLY [3 votes]: By a standard argument that also applies in this case, polynomially parallel many witness queries give the same power as $O(\log n)$ many sequential witness queries:
$$\mathrm{FP^{\|\Sigma^P_1[wit]}=FP^{\Sigma^P_1[wit,\log]}}.$$
In particular, the predicates (i.e., 0–1 valued functions) in $\mathrm{FP^{\Sigma^P_1[wit]}}$ are exactly those from $\Delta^P_2=\mathrm{P^{NP}}$, whereas predicates in $\mathrm{FP^{\|\Sigma^P_1[wit]}}$ are exactly those from the complexity class $\Theta^P_2=\mathrm{P^{NP[\log]}=P^{\|NP}=L^{NP}}$ (see zoo). It is widely expected that $\Theta^P_2\ne\Delta^P_2$, and this is supported by oracle separation results. Thus, the answer to your question should be no.
See also https://cstheory.stackexchange.com/q/31482 for more information about this class.