 Welcome back to our lecture series math 42-30 abstract algebra 2 for students at Southern Utah University. As usual, I'll be a professor today, Dr. Angela Seline. Alas, we come to the last episode in this lecture series, lecture 39, for which we're going to continue developing our notion of a Boolean algebra. In this video, I want to prove the so-called De Morgan laws of Boolean algebra. These are properties we might know from logic or we might know from set theory, but as those are both examples of Boolean algebras, this is actually a property that extends to the general Boolean algebra setting. So imagine B is a Boolean algebra and let X and Y be elements of said Boolean algebra. Then the following is true. If we take X join Y complement, that is we complement the whole thing. This is equal to X complement, meet Y complement. And similarly, if we take X meet Y and the complement of that whole thing, that is equal to X complement join Y complement. In particular, when you take complements of an operation of a Boolean algebra, you are going to take the complements of the individual elements, but the operation will switch from meet to join and or excuse me, from join to meet and meet to join in this situation. Now, because of the principle of duality, it suffices for us to prove one of these statements. We only have to prove the first one that X join Y complement is equal to X complement join Y complement because duality will take care of the other one right here. Now, how are you going to do this? What we're going to do is because previously we've shown that in a Boolean algebra, complements are unique. There's only one complement given an element. So the complement for X join Y is uniquely determined. So therefore, if X complement meet Y complement, walks like a complement, and quacks like a complement, that makes it the complement of X join Y. So that's exactly what we're going to do. We're going to take X join Y and we're going to join it with X complement meet Y complement. So if these are in fact complements, an element join its complement, that should equal the maximum element one. And then down below, we're going to show that X join Y meet X complement meet Y that this is going to equal zero in that situation because that's exactly what complements are going to do. So we're going to prove that this element acts like the complement. And like I said, the second equality follows by duality. So let's start with the first one here. X join Y join X complement meet Y complement. We want to show that this thing is equal to one. Okay, so what can we do with this thing? Well, because we have a join and a join, we can use associativity and we can rewrite this as X join Y join X complement meet Y complement like so. For which then the next thing is since we have a join and a meet, we can distribute the join Y across the meet. And so this is going to give us Y join X complement meet Y join Y complement. But wait a second, we know that Y complement is the complement of Y. Clearly it's in the name. So Y join Y complement that's equal to one. And remember, one is the identity for the meet operation. So one meet anything is always that thing. So this part is just going to just vanish off. And we're left with now X join Y join X complement like so. All right, so now we are commutative. So we can swap the order of those. So becomes X join X complement join Y join is associative. So we can rewrite this as X join X complement join Y. And just like above, since X and X complement are complements of each other, X join X complement is equal to one. For which as one is the dominant element of join one join anything is always equal to one. So we get a one right here. And this proves the first statement that if you take X join Y and you join it with its potential complement, you actually get one, which is what you should get for a complement. Now that's not enough. We have to go the other direction. We have to show that X join Y when you meet it with its, you know, its supposed complement, this candidate that that's going to equal zero. And so we're going to do that right here now. This one's a little bit shorter, right? So to consider this, we're going to take X join Y meet X complement join Y complement. Well, the first thing we're going to do is that meet is commutative. So swap the order like so. Okay, then the next thing we're going to do is use the fact that the double complement is the original element. We prove previously that X double complement is equal to X and same things can be said for Y. So X is equal to X double complement and Y is equal to Y double complement like so. And so I want to remind us about the statement we had just proven, right? We have just proven that A join B join A complement meet B complement is equal to one. So that's a statement we had just proven. It's not along on the screen here. Because this statement is true, its dual statement has to likewise be true. The dual statement would be A meet B meet. In this case, A complement join B complement. And that's going to equal zero in that situation. Now if you use the substitution that A equals X complement and B equals Y complement, we're exactly in this situation right here, the left hand side, we have X complement meet Y complement meet X double complement join Y double complement. That statement by the dual of the previous one is thus equal to zero like so. And so that then gives us that direction much, much shorter we'll benefit from the statement we had before. So this shows that X complement join excuse me, X complement meet Y complement is in fact a complement for X join Y as complements are unique. That tells us that this must be the complement of X join Y. So we complete the statement right here. And then like I mentioned, the second statement is actually the dual of the first statement. And so that then proves the De Morgan laws for any Boolean algebra.