Are there incompatible Turing degrees $a,b$ s.t any degree computable in $a$ either computes $a$ or is computed by $b$?


Obviously, if $a$ was above $b$ then $a$ would be a strong minimal cover of $b$. But do incomparable such degrees exist and is there a name for them?


Or am I missing some obvious fact that implies such degrees can't exist? I feel like I should already know the answer but i don't seem to.


Edit: I meant to rule out the case where this is trivially possible so I should have added: and this relation holds for no $b$ below $a$.



Incompatible degrees $a,b$ s.t. $x < a$ implies $x \leq b$
This type of algebraic property is observed in the local structure of the enumeration degrees: there we call them Ahmad pairs. Ahmad (a student of Lachlan) showed that there are incomparable $\Sigma^0\_2$ enumeration degrees $a$ and $b$ such that for all enumeration degrees $x$ if $x<a$ then $x<b$. She also showed that there are no symmetric Ahmad pairs. This property plays a key role in our attempts to analyze the 2-quantifier theory of the structure.


In the c.e. Turing degrees there are no Ahmad pairs because of Sacks splitting theorem (in an Ahmad pair the first element cannot be the join of two lower ones). In the $\Delta^0\_2$ Turing degrees there are symmetric Ahmad pairs: any pair of distinct minimal degrees.


