A detail in the proof of Stone representation Theorem Let $(\mathcal{B},\sqcap,\sqcup,\leq)$ be a Boolean algebra. Let $x,y\in\mathcal{B}$. I want to prove the following implication:
$$x\sqcap y'\leq 0\Rightarrow  x\leq y$$
where $y'$ is the complement of $y$.

I have checked that this works in the case $\mathcal{B}=\mathcal{P}(X)$, with $\sqcap$=intersection, $\sqcup$=union, $\leq=\subseteq$, $0=\varnothing$.....but I can't prove it for a generic Boolean algebra.
Hint: $x=x\sqcap1=x\sqcap(y\sqcup y')=(x\sqcap y)\sqcup(x\sqcap y')=\ldots$
Since $$x⊓y′≤0$$
$$(x⊓y′)⊔y ≤0 ⊔ y = y$$
$$(x⊔y)⊓(y'⊔y)≤ y$$
$$(x⊔y)⊓1 ≤ y$$
$$(x⊔y)≤ y$$
$$x ≤ y$$