How to prove boolean ordering question

Question

Let $\sqsubseteq$ be the boolean ordering of $X$, so for every $x$ and $y$ 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 that $v \sqcup w \sqsubseteq a\sqcup b$ and $v \sqcap w \sqsubseteq a\sqcap b$.

Should this be solved algebraically, or in a different way? And if so, where would be my starting point?

Answer 1

HINT: Work up to it in steps:

1. Show that if $x\sqsubseteq y$, then $x\sqcap a\sqsubseteq y\sqcap a$.
2. Show that if $x\sqsubseteq y$, then $x\sqcup a\sqsubseteq y\sqcup a$; you may find it helpful to show first that $x\sqsubseteq y$ iff $x\sqcup y=y$.
3. Apply (1) and (2) twice to get the desired results.

Answer 2

Hint:

• What you want to prove is that $\sqcup$ and $\sqcap$ are monotone with respect to $\sqsubseteq$.
• Start with a simple example: prove that the ordinary set operations $\cup$ and $\cap$ are monotone with respect to subset relation $\subseteq$, i.e.

\begin{align} A_1 \subseteq A_2 \land B_1 \subseteq B_2\quad &\implies\quad A_1 \cup B_1 \subseteq A_2 \cup B_2 \\ A_1 \subseteq A_2 \land B_1 \subseteq B_2\quad &\implies\quad A_1 \cap B_1 \subseteq A_2 \cap B_2 \end{align}

• Transform your proof so that is uses only set-operations and related theorems and axioms.
• Try to make the proof work for an arbitrary boolean algebra.

I hope this helps $\ddot\smile$