Prove that: if $x \sqcup \bar{y}=1$, then $x \sqcup y=x$ (in a Boolean algebra)

Question

Suppose X is a Boolean algebra. Prove that: if $x \sqcup \bar{y}=1$, then $x \sqcup y=x$

I suspect this one is not that difficult, but for some reason I can't find the answer. This homework assignment question even has a hint: "Start with $x \sqcup y$ and use a property of 1."

Any suggestions or hints anyone?

Answer 1

See Boolean algebra.

We have that :

$x = x \sqcup 0 = x \sqcup (y \sqcap \overline y) = (x \sqcup y) \sqcap (x \sqcup \overline y)$.

Now, if $x \sqcup \overline y = 1$, we have :

$x = (x \sqcup y) \sqcap (x \sqcup \overline y) = (x \sqcup y) \sqcap 1 = x \sqcup y$.

Answer 2

I believe ⊔ stands for "OR". Assuming $\cap$ denotes "AND", we have $x\cup y = (x\cup y)\cap (x\cup {\bar y})$ (using the property of 1). Using the distributive property of $\cup$ over $\cap$, we have $x\cup y = (x\cup y)\cap (x\cup {\bar y}) = x\cup (y \cap {\bar y}) = x\cup 0 = x$ (because $x$ "OR" $0$ is always $x$.)

In case, you are not familiar with distributive law: $A\cup (B\cap C) = (A\cup B)\cap(A\cup C)$. One can interchange the $\cup$ and $\cap$ and the identity is still valid in Bolean algebra.

Answer 3

just write out all the possibilities for x & y. There are only 4 combinations possible...