Can this puzzle be solved mathematically?

Proof that $4$ is a lower bound for the number of shifts required:

There are $6\cdot5=30$ ordered pairs $(A,B)$ of distinct people from the people in the ship, so the shifts must combine to have at least $30$ ordered pairs of (awake person, asleep person).

If in a given shift $k$ people are awake and $6-k$ are asleep, then the number of such ordered pairs is $k(6-k)$. This is maximized when $k=3$, when there are $9$ such ordered pairs.

Therefore it is impossible for $3$ shifts to cover all possibilities, since there can be at most $27$ ordered pairs covered between them.