# Let $A=(A,\vee,\wedge)$ be a Boolean Algebra. So $\overline{x}=y$ is enough that...?

Let $$A=(A,+,\cdot)=(A,\vee,\wedge)$$ be a Boolean Algebra. To ensure that $$\overline{x}=y$$ is sufficient that:

1. $$x+y=1_A$$.
2. $$x\cdot y=0_A\;\wedge\;x\neq0_A$$.
3. $$x\cdot(x+y)=\overline{y}\cdot x+\overline{(y+x)}$$
4. None of the above.

I have some doubts.

# First question

First of all, do we have to find $$\text{??}$$ in $$\forall x,y\in A(\text{??}\to\overline{x}=y)$$?

# Second question

I think the correct answer is (3) because: \begin{align*} x\cdot(x+y)=\overline{y}\cdot x+\overline{(y+x)}&\implies x\wedge(x\vee y)=(\overline{y}\wedge x)\vee\overline{(y\vee x)}\\ &\implies x=(\overline{y}\wedge x)\vee(\overline{y}\wedge\overline{x})\\ &\implies x=\overline{y}\wedge(x\vee\overline{x})\\ &\implies x=\overline{y}\wedge1_A=\overline{y}\\ &\implies\overline{x}=\overline{\overline{y}}\implies\boxed{\overline{x}=y} \end{align*}

but I am not sure if I also have to discard the other possibilites (because we could find 2,3,... sufficient conditions, not only 1).

# Third question

Can you help me finding counterexamples for (1), (2) and (4)?

For example for (1) I said $$(D_6,\mid)$$ where $$D_6=\text{Divisors of 6}=\{1,2,3,6\}$$, which is a Boolean Algebra, and its Hasse diagram is:

but $$\forall x,y\in A(x\vee y=1_A\to\overline{x}=y)$$ is false. Indeed, take $$x=2$$ and $$y=6$$. Then $$2\vee6=6$$ but $$\overline{2}=3\neq6$$.

What about (2) and (4) (if my counterexample is correct)?

For example, $$(D_{30},|)$$. Here, $$12\vee 15=30$$, which is the top, but $$12\wedge15=3$$ which is not the bottom, and so (1) is not sufficient.
Check that also $$2\wedge3$$ is the bottom, but $$2\vee3$$ is not the top, so that $$2$$ is not the complement of $$3$$; so (2) is not sufficient either.