every lattice satisfies $x\leq y \rightarrow x\vee(y\wedge z) \leq y\wedge(x\vee z)$.
My proof:
$x\leq y, x\leq (x\vee z)\Rightarrow x\leq y\wedge(x\vee z)$
$(y\wedge z)\leq y,z$ and $x,z\leq (x\vee z) \Rightarrow (y\wedge z)\leq (x\vee z)$
$(y\wedge z)\leq y, (y\wedge z)\leq (x\vee z)\Rightarrow (y\wedge z)\leq y\wedge(x\vee z)$
$x\vee(y\wedge z) \leq y\wedge(x\vee z)$
is this proof correct or it's not clear and I missed something ?
Your proof is Ok, but it's actually simpler (or cleaner, in my opinion) with the following approach.
If $x \leq y$ then $$x \vee (y \wedge z) \leq y \vee (y \wedge z) = y.$$ I suppose you're familiar with absorption, which is valid in all lattices: $$x \vee (x \wedge y)=x \quad\text{and}\quad x \wedge (x\vee y)=x.$$ Now we need to show that $x\vee (y \wedge z) \leq x \vee z$.
But since $y \wedge z \leq z$, $$x \vee (y\wedge z)\leq x\vee z.$$