Is there a quick trick to see $\mathbb{E}[\theta|L=l] = \mathbb{E}_{\Lambda}\left[ \mathbb{E}[\theta|\Lambda, L=l] | L=l \right]$ without expanding?

Question

As in the title, why is $\mathbb{E}[\theta|L=l] = \mathbb{E}_{\Lambda}\left[ \mathbb{E}[\theta|\Lambda, L=l] | L=l \right]$?

A quick glance says tower rule, but not so obvious how is it applied here.

EDIT: $\mathbb{E}_{\Lambda}$ just means that the expectation is taken wrt to $\Lambda$ (uses the density of $\Lambda$). Further, if you expand these expectations (as in, write everything in terms of integrals), then everything works out, they are equal. But the equality was given as a one-liner (in an otherwise spoon-fed class) to me, that makes me think there is a quick trick/ property that was used here.

Answer 1

By the tower rule: $$\mathbb{E}[\theta\mid L] =\mathbb{E}\Big[\mathbb{E}[\theta\mid\Lambda, L]\mid L\Big]$$ From where: $$\mathbb{E}[\theta\mid L=l] = \mathbb{E}\Big[ \mathbb{E}[\theta\mid\Lambda, L]\mid L=l \Big]=\mathbb{E}\Big[ \mathbb{E}[\theta\mid\Lambda, L=l]\mid L=l \Big]=$$ $$=\mathbb{E}_{\Lambda}\Big[ \mathbb{E}[\theta\mid\Lambda, L=l]\mid L=l \Big]$$