TITLE: Are there two computable binary trees such that each has a branch not computing any branch through the other?
QUESTION [9 upvotes]: It is a well-known elementary classical result in computability theory that there are computable infinite binary trees $T\subset 2^{<\omega}$ having no computable infinite branch. (One can build such a tree $T$ as follows: fix a computably inseparable pair of c.e. sets $A$ and $B$, and allow a binary string $t$ into the tree when the pattern of yes/no answers is consistent with being a separation of $A$ and $B$ in as much as these sets are revealed by stage $|t|$. In this way, incorrect guesses about the separation are eventually killed off, stuck in a finite dead part of the tree, and there can be no computable infinite branch because the sets have no computable separation.)
What I have need of is a strengthening of that classical result to the following:
Question. Are there two computable trees $T,S\subset 2^{\lt\omega}$ such that $T$ has an infinite branch not computing any infinite branch through $S$ and $S$ has a branch not computing any branch through $T$? 
That is, I want computable trees $S$ and $T$ such that there are infinite binary sequences $s$ and $t$, which are branches through $S$ and $T$, respectively, such that no infinite binary sequence that is Turing computable from $s$ is a branch through $T$ and no infinite binary sequence that is Turing computable from $t$ is a branch through $S$. (So in particular, neither $S$ nor $T$ can have any computable infinite branches.)

REPLY [2 votes]: Here is another example.
Let $P_1$ be a nonempty $\Pi^0_1$-class which only contains Martin-Lof random reals, and $P_2$ be a nonempty $\Pi^0_1$-class in which no path is either of DNR degree or recursive. The existence of $P_2$ can be found in Rod and Denis's book (Thm 12.4.6, page 584).
Now it is obvious no real in $P_2$ computing a real in $P_1$. Let $x$ be a hif random real, then there must be some $y\in P_1$ Turing equivalent to $x$. However, any nonrecursive real Turing below $x$ must be DNR. So no real in $P_2$ can be computed by $y$.