We have a poset $(X, \sqsubseteq)$, and we define operations $+$ and $\cdot$ by $x+y=inf(x, y)$ and $x\cdot y=sup(x, y)$ ($+$ can be seen as union in sets and $\cdot$ as intersection in sets).
The question is: show that $+$ and $\cdot$ satisfy the absorption property $x+(x\cdot y)=x$. There is a suggestion to first settle $\textit 0$ and $\textit 1$ and also make use of symmetry.
It has the following Hasse diagram:
My idea is that I should translate $x+(x\cdot y)=x$ to $inf$ and $sup$, like this:
$sup(1, inf(1, 0))=\\sup(1, 0)=\\1$
But to me, that feels like not proving that much. Also, I didn't make use of symmetry as was suggested.
How can I tackle this?
I prefer to denote the operations by $x\land y=\inf(x,y)$ and $x\lor y=\sup(x,y)$, because $+$ and $\cdot$ might suggest wrong ideas.
Those operations are well defined because your poset is a lattice. Now, by definition, $x\land y\sqsubseteq x$, so $x$ is an upper bound of the subset $\{x,x\land y\}$. Can the least upper bound be different from $x$?
When you have realized an answer to the question, you'll have proved that $x\lor(x\land y)=x$.
You can also do this by cases; prove the property for $x=0$, for $x=1$ and for the case when neither $x$ or $y$ is $0$ nor $1$. And for this it's sufficient to check $x=p$ and $y=q$, as permuting the names of $p$, $q$ and $r$ doesn't change the Hasse diagram (in a substantial way).