TITLE: Does forcing with recursively pointed perfect trees add a Turing degree that is minimal over $V$?
QUESTION [8 upvotes]: A tree $T$ on $\omega$ is recursively pointed if it is recursive in each of its branches.  We can consider a variant of Sacks forcing where the conditions are recursively pointed perfect trees ordered by inclusion.  Given a $V$-generic filter $G$ for this forcing, we can define the real $x_G = \bigcap_{T\in G} [T]$. 
For every real $x \in V$ we have $x <_\text{T} x_G$.  Given a real $x \in V[G]$ with $x <_\text{T} x_G$, must we have $x \in V$?
Probably the answer to this question is well-known, but I'm afraid I don't see how the proof of the minimality property for Sacks forcing (as given by Jech) might adapt to recursively pointed trees.

REPLY [7 votes]: If you two haven't worked this out already, it seems to me that $x_G$ is not minimal as a Turing degree above $V$, but is a minimal $V$ degree.  For the first, since $x_G$ computes $0'$, it follows that $x_G$ is the join of two mutually 1-generic reals, neither of which computes $x_G$ and at least one of which is not in $V$.  However, if $x_G$ computes $x$ and $x$ is not in $V$, then there is an $a$ in $V$ such that $x\oplus a$ computes $x_G$.  This is by the usual minimality argument: $a$ is the degree of the splitting tree for the functional used to compute $x$ from $x_G$.