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.