If A and B are nonzero c.e. degrees and the intersection of A and B is computable, must A and B be a minimal pair?

Question

Furthermore, does the converse hold?

By minimal pair I mean a pair (A,B) such that if some language C turing reduces to A and C turing reduces to B, then C is computable.

Answer 1

I'm assuming that $A$ and $B$ should be sets, not degrees; otherwise per my comment the question doesn't make much sense.

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Set-theoretic operations don't generally tell you much about reducibility and degrees. For example, let $X$ and $Y$ be any two sets and consider $EX=\{2x: x\in X\}$ and $OY=\{2y+1: y\in Y\}$. Then $EX\equiv_TX, OY\equiv_TY$, and $EX\cap OY=\emptyset$. So the set-theoretic intersection tells us nothing.

Similarly, given any two sets $X$ and $Y$ the sets $\{2z: z\in X\}\cup\{2z+1: z\in\mathbb{N}\}$ and $\{2z+1: z\in Y\}\cup\{2z: z\in\mathbb{N}\}$ are Turing equivalent to $X$ and $Y$ respectively but their union is everything, so the union tells us nothing.

In particular, the answer to both questions you ask is no.

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The situation is somewhat more interesting with the symmetric difference. Here things are more subtle: sometimes symmetric difference is well-defined on degrees. A bit about this is said here; Uri Andrews, Peter Gerdes, Joe Miller, Mariya Soskova, and I worked a bit more on the topic, but that hasn't been written up yet.