Let $R$ be a ring with unity. For any $a\in R$, prove that $$l_R(r_R(l_R(a)))=l_R(a),$$ where $l_R(a)$ is the left annihilator of $a$ and $r_R(a)$ is the right annihilator of $a$.
My attempt: Let $y\in r_R(l_R(a))$ such that $(l_R(a))y=0$. Then $l_R(a)\subseteq l_R(y)=l_R(r_R(l_R(a)))$ which proves $l_R(a)\subseteq l_R(r_R(l_R(a)))$. However, I have failed to figure out the reverse inclusion.
It follows immediately from the containment reversing properties of both maps, and the “extending property” of $rl$ and $lr $. This is straightforward to prove, and less complex than verifying the equalities you are doing directly.
The properties I’m talking about are: if $a\subseteq b$, then $l(b)\subseteq l(a)$ and $r(b)\subseteq r(a)$. Further, $rl(a)\supseteq (a)$ and $lr(a)\supseteq (a)$.
From there it follows that $rl(a) \supseteq a$, and then $lrl (a)\subseteq l(a)$.
It also follows that $lr(l(a))\supseteq l(a)$.