They are trying to prove that
any rectangular grid with side-lengths congruent to each other and not to zero modulo 3, with a single cell removed in the middle, can be covered by L-trominoes.
How are they doing it?
By induction: starting with the $(3M\pm1)\times(3N\pm1)$ rectangle shown on the left, reducing it to the cases of four smaller rectangles, and continuing to apply the same reduction multiple times until they end up with a $2\times(3z-1)$ rectangle, which can be covered as shown on the right.
Will they succeed?
Unknown. The proof doesn't guarantee that they'll always end up with a collection of $2\times(3z-1)$ rectangles without involving any $1\times(3z+1)$ rectangles (which clearly aren't coverable by trominoes). In fact, it appears that the general statement being considered is an open problem.
A special case of the result being considered may be found
at Maths SE (spoilers, obviously). The L-ish proof here uses the same underlying idea, but with the difference that
we can't always split the board into four equally sized rectangles each time.
reverse-puzzling
in lieu of aproof-without-words
category $\endgroup$