[proofplan] We use only the universal properties of meet and join in a lattice. Idempotence, commutativity, and associativity follow by showing that the two displayed expressions have the same greatest-lower-bound or least-upper-bound property. The absorption identities are proved by identifying $x$ as the greatest lower bound of $x$ and $x\vee y$, and as the least upper bound of $x$ and $x\wedge y$. Finally, the order relation is recovered from the defining inequalities of meet and join. [/proofplan]

[step:Derive idempotence and commutativity from uniqueness of bounds] Fix $x,y\in L$. Since $x$ is a lower bound of the pair $\{x,x\}$, and every lower bound of $\{x,x\}$ is below $x$, the element $x$ is the greatest lower bound of $\{x,x\}$. Hence $x\wedge x=x$. The same argument with upper bounds shows that $x$ is the least upper bound of $\{x,x\}$, so $x\vee x=x$. Both $x\wedge y$ and $y\wedge x$ are greatest lower bounds of the same two-element set $\{x,y\}$: each is below both $x$ and $y$, and every common lower bound of $x$ and $y$ is below each of them. By antisymmetry of $\le$, $x\wedge y=y\wedge x$. Similarly, both $x\vee y$ and $y\vee x$ are least upper bounds of $\{x,y\}$, so $x\vee y=y\vee x$. [/step]

[step:Identify both bracketings as the same triple meet and triple join] Fix $x,y,z\in L$. Let $m_1=(x\wedge y)\wedge z$. Then $m_1\le x\wedge y$, so $m_1\le x$ and $m_1\le y$, and also $m_1\le z$. Thus $m_1$ is a lower bound of $\{x,y,z\}$. If $u\in L$ satisfies $u\le x$, $u\le y$, and $u\le z$, then $u\le x\wedge y$, and therefore $u\le (x\wedge y)\wedge z=m_1$. Hence $m_1$ is the greatest lower bound of $\{x,y,z\}$. Let $m_2=x\wedge (y\wedge z)$. The same verification gives that $m_2$ is also the greatest lower bound of $\{x,y,z\}$: it is below $x$, $y$, and $z$, and every common lower bound of these three elements is below $y\wedge z$ and hence below $m_2$. By uniqueness of greatest lower bounds, $m_1=m_2$, so \begin{align*} (x\wedge y)\wedge z=x\wedge (y\wedge z). \end{align*} For joins, let $j_1=(x\vee y)\vee z$. Then $x\le j_1$, $y\le j_1$, and $z\le j_1$. If $v\in L$ satisfies $x\le v$, $y\le v$, and $z\le v$, then $x\vee y\le v$, and hence $j_1\le v$. Thus $j_1$ is the least upper bound of $\{x,y,z\}$. Let $j_2=x\vee (y\vee z)$. Again, $j_2$ is an upper bound of $\{x,y,z\}$, and every common upper bound of $x$, $y$, and $z$ lies above $j_2$. Hence $j_2$ is also the least upper bound of $\{x,y,z\}$. By uniqueness of least upper bounds, \begin{align*} (x\vee y)\vee z=x\vee (y\vee z). \end{align*} [/step]

[step:Prove the absorption identities by checking their universal properties]Fix $x,y\in L$. Since $x\le x$ and $x\le x\vee y$, the element $x$ is a lower bound of $\{x,x\vee y\}$. If $u\in L$ is any lower bound of $\{x,x\vee y\}$, then in particular $u\le x$. Hence $x$ is the greatest lower bound of $\{x,x\vee y\}$, and therefore \begin{align*} x\wedge (x\vee y)=x. \end{align*} Also, since $x\le x$ and $x\wedge y\le x$, the element $x$ is an upper bound of $\{x,x\wedge y\}$. If $v\in L$ is any upper bound of $\{x,x\wedge y\}$, then in particular $x\le v$. Hence $x$ is the least upper bound of $\{x,x\wedge y\}$, and therefore \begin{align*} x\vee (x\wedge y)=x. \end{align*}[/step]

[guided]We prove the two absorption identities by returning directly to the definitions of meet and join. First consider $x\wedge (x\vee y)$. By definition, this element is the greatest lower bound of the two elements $x$ and $x\vee y$. To prove that it equals $x$, we prove that $x$ itself has exactly that greatest-lower-bound property. The element $x$ is a lower bound of $\{x,x\vee y\}$ because $x\le x$ by reflexivity and $x\le x\vee y$ by the defining property of the join $x\vee y$. Now let $u\in L$ be any lower bound of $\{x,x\vee y\}$. Being a lower bound means that $u\le x$ and $u\le x\vee y$. In particular, $u\le x$. Thus every lower bound of $\{x,x\vee y\}$ lies below $x$, while $x$ itself is a lower bound. Therefore $x$ is the greatest lower bound of $\{x,x\vee y\}$. Since meets are greatest lower bounds, we obtain \begin{align*} x\wedge (x\vee y)=x. \end{align*} Now consider $x\vee (x\wedge y)$. This element is the least upper bound of $x$ and $x\wedge y$. We prove that $x$ itself has this least-upper-bound property. The element $x$ is an upper bound of $\{x,x\wedge y\}$ because $x\le x$ by reflexivity and $x\wedge y\le x$ by the defining property of the meet $x\wedge y$. If $v\in L$ is any upper bound of $\{x,x\wedge y\}$, then $x\le v$ and $x\wedge y\le v$. In particular, $x\le v$. Thus $x$ is below every upper bound of the pair, while $x$ itself is an upper bound. Therefore $x$ is the least upper bound of $\{x,x\wedge y\}$, so \begin{align*} x\vee (x\wedge y)=x. \end{align*}[/guided]

[step:Recover the order relation from meet and join] Fix $x,y\in L$. Suppose first that $x\le y$. Then $x$ is a lower bound of $\{x,y\}$. If $u\in L$ is any lower bound of $\{x,y\}$, then $u\le x$. Hence $x$ is the greatest lower bound of $\{x,y\}$, so $x\wedge y=x$. Also, $y$ is an upper bound of $\{x,y\}$. If $v\in L$ is any upper bound of $\{x,y\}$, then $y\le v$. Hence $y$ is the least upper bound of $\{x,y\}$, so $x\vee y=y$. Conversely, if $x\wedge y=x$, then the defining inequality $x\wedge y\le y$ gives $x\le y$. If $x\vee y=y$, then the defining inequality $x\le x\vee y$ gives $x\le y$. Therefore \begin{align*} x\le y \quad \Longleftrightarrow \quad x\wedge y=x \quad \Longleftrightarrow \quad x\vee y=y. \end{align*} This completes the proof. [/step]