Let $\sqsubseteq$ be a boolean ordering of the boolean algebra $X$, which means that for each $x$ and $y$ the following applies: $x \sqsubseteq y$ if $x \sqcap y = x$.
Let $v, w, a, b \in X$ with $v \sqsubseteq a$ and $w \sqsubseteq b$.
Show/prove the following:
that
$v \sqcup w \sqsubseteq a \sqcup b$
and that
$v \sqcap w \sqsubseteq a \sqcap b$
Now, I can answer this question with the help of set theory:
if $V \subset A$, then it follows that $V \subset A\cup B$. Also, from $W \subset B$ it follows that $W \subset A\cup B$.
It then follows that if both $V$ and $W$ are subsets of $A\cup B$, it also follows that $V\cup W \subset A \cup B$.
The same story goes for $V \subset A$ and $V \subset A \cap B$.
But how can I show/express this in boolean algebra notation? How can I use $x \sqsubseteq y$ if $x \sqcap y = x$ to prove $v \sqcup w \sqsubseteq a \sqcup b$ and $v \sqcap w \sqsubseteq a \sqcap b$?