# Complements being pseudocomplements in a lattice

I'm reading an article and it says that, if $$l$$ has a $$\bar{l}$$ complement in a distributive lattice $$L$$, then $$\bar{l}$$ is a pseudocomplement of $$l$$. Can someone point me a hint, I'm new with lattice theory.

$$\bar{l}$$ is complement of $$l$$ if $$l\vee\bar{l}=\top$$ and $$l\wedge\bar{l}=\bot$$.

A pseudocomplement of $$l\in L$$ is an element $$l^{\star}$$ such that $$m\leq l^{\star}$$ if and only if $$m\wedge l=\bot$$.

If $$m \le \bar{l}$$ then $$m \land l \le \bar{l} \land l = \bot$$ is immediate (in any lattice $$x \le y$$ implies $$x \land z \le y \land z$$ for any $$x,y,z$$). And always $$\bot \le m \land l$$ so $$m \land l = \bot$$.
OTOH, if $$m \land l = \bot$$, then $$\bar{l}= \bar{l} \lor \bot = \bar{l} \lor (m \land l)$$ and we apply distributivity to get $$\bar{l}= (\bar{l} \lor m) \land (\bar{l} \lor l) = (\bar{l} \lor m) \land \top = \bar{l} \lor m$$ so in conclusion
$$\bar{l} = \bar{l} \lor m \implies m \le \bar{l}$$ as required.