# Showing that an algebraic lattice is a lattice.

Let $$(L, \wedge, \vee)$$ be an algebraic lattice. If we define $$\begin{equation*} x \le y :\iff x \wedge y = x \end{equation*}$$ then $$(L,\le)$$ is a lattice ordered set.

This is what I have to prove (without using duality). First I showed that $$(L,\le)$$ is a poset proving partial order relations. Then I showed that $$\sup L=x\vee y$$. But I am not able to show that $$\inf L=x\wedge y$$ as the relation defined in the question is for $$\wedge$$.

So if I want to show that $$\inf L=x\wedge y$$, firstly I need to show that it is a lower bound. For that, I should use Absorptive laws: $$x\vee(x\wedge y)=x\not\implies x\ge x\wedge y$$ and $$y\vee(y\wedge x)=y\not\implies y\ge (y\wedge x)=(x\wedge y)$$ which shows nothing for $$(x\wedge y)$$ to be a lower bound for the pair $$x,y\in L$$. Which argument can I use? Because the condition says that $$x\le y\iff x\wedge y=x$$. Can I redefine the condition as $$y\ge x\iff y\vee x=y$$ (something like this)?

You can deduce $$x \leq y \iff x \lor y = y$$ as follows:

\begin{align*} x \leq y &\Rightarrow x \land y = x \\ &\Rightarrow (x \land y) \lor y = x \lor y &&\text{ we just join y at both sides}\\ &\Rightarrow y = x \lor y &&\text{ by absorption law} \end{align*}

In the other direction:

\begin{align*} y = x \lor y &\Rightarrow y \land x = (x \lor y) \land x \\ &\Rightarrow y \land x = x \\ &\Rightarrow y \geq x \end{align*} Using again just absorption and the definition given by the question. The rest, as you already noticed, should follow from that.

edit: correction pointed out in the comments.

• You are right. My mistake. Commented Apr 6, 2021 at 20:48
• can't we just join $y$ from both sides? Commented Apr 6, 2021 at 20:53
• But this was not my question. My question was how I show that $x\wedge y$ is a lower bound. Commented Apr 6, 2021 at 20:55
• Well, you asked if you could re-define $x \leq y \iff x \lor y = y$. I assumed by your phrasing you knew how to go from there to solve your question. Commented Apr 6, 2021 at 20:56
• actually there are multiple questions. As I mentioned in my question that the defining operator is $\wedge$. With this, how can I show $\wedge$ is lower bound? Though +1 for your answer. Commented Apr 6, 2021 at 20:59