 In the previous video, we have defined the notion of a Boolean algebra. In this video, I want to collect some properties about Boolean algebras. Now, this discussion is going to mimic that is our previous discussions of properties of groups, rings, fields, vector spaces, r modules, you know, it's very similar things. So Boolean algebras do have identities zero and one. And as we already mentioned, when we talked about bounded lattices, these identities are necessarily unique to each other. And this is a consequence of just, you know, we have a two-sided identity. I want to prove that in a Boolean algebra, we have cancellation laws. So for example, our cancellation laws look like the following. If x join y is equal to x join z and x complement join y is equal to x complement join z, then that actually means that y equals z. Okay. Similarly, if x meet y is equal to x meet z and x complement meet y is equal to x complement meet z, then you can cancel the x's and you get that y equals z in that situation. And then the third cancellation law says that if x join y equals x join z, and if x meet y is equal to x meet z, then in that situation, you get that y equals z. Now, when it came to groups, all we had to say was something like the following. If x times y equals x times z, then you can cancel the x and you have y equals z. Now, of course, in groups, you have inverses, so we could use inverses to cancel things out. We don't have inverses. We have compliments, but that's what will be required here, that if two products are the same with regard to x and with regard to its x complement, then it turns out the factors had to be the same. And that doesn't matter if you're doing joins or meets. In this situation, this says that if x acts the same way with respect to join as it does meet, then you can cancel the x out and get that y equals z in that situation. So there's a couple of things we have to prove here. Now, before going on, I do want to mention that with the groups like this is only one sided. You didn't have a notion of a complement there. You do need both statements in order for these things to be true. So note that cancellation is only going to work with both hypotheses because if you take something like a following, let's take z is less than y, which is less than x, and these are strict inequalities here. If you take that situation, since x is greater than y, notice this is going to equal x. x join y is going to equal x. And then because x is greater than z, you're going to get that x join z is likewise equal to x. So these things are equal to each other. But since z and y are not equal to each other, we can't infer that y equals z because that would actually be a falsehood. So we do have to consider the complement. Similar kind of examples can be constructed for these other two possibilities as well. So what we're going to do is we're going to prove the first one. So let's assume that we have equality when you join the x and when you join x complement. We want to show that y equals z in that situation. So let's start off with just y. y equals zero join y because zero is the join identity. Well, since this is the join identity, we can use complements. The join identity zero is equal to x meet x complement. Again, it's kind of like an inverse, but it's backwards in this situation. The complements combined to give you the dominant element, not the identity element. So zero is equal to x join x complement, excuse me, x meet x complement join y like so. For which now we're going to distribute join across meet. So this becomes x join y meet x complement join y, but by our assumptions x join y is x join z and x prime join y is the same thing as x prime join z. For which then, since we have z in the left and the right factor, we can factor it out by the distributive law. So we get x join x prime meet z, excuse me, x meet x prime join z in that situation. Well, x meet x prime, since they're complements, that's going to give you zero. Zero join z gives you back z, and so that proves the first cancellation law that was then here. So you do need the complements. So we took care of that one. Now the second one, I don't need to provide a proof for it because the second one is the dual statement of the first one. So take the proof, we just saw, take its dual and boom, that's the proof of the second one. Now the third one here is actually a so-called self-dual statement. If you reverse all the symbols, meet with join and join with meet, and you'd also switch things like inequalities and zeros and ones, but that's not in the statement. If you switch all the joins with meet to meet with joins, you get the exact same statement back. So it's self-duals. Now since this is a self-dual statement, we have to prove it because it's not the dual of anything else other than itself. So we'll provide it. It'll be a very similar proof to what we just saw. So let's come down and take a look at what that looks like. So we want to prove that y equals z. So we're going to start off with y and then end up with z. So what we have here is that y, of course, is the element we're starting with. And so then I claim that this equals y, join, x, meet, y. And this is a consequence of the absorption property there. I also used some commutativity going on there because absorption is like y, join, y, meet, x is equal to y, and then I swap the order of these things. Took some liberty there. But absorption commutativity gives us this one right here. Then by assumption, x, meet, y is equal to x, meet, z. So I make that substitution right there. The next thing we're going to do is then we're going to distribute y in this situation. So this is going to then give us y, join, x, meet, y, join, z, for which in the second case by commutativity we can swap these things around. So we get the statement now that x, join, y, meet, y, join, z is where we act now. But then by assumption from before, x, join, y is the same thing as x, join, z. And then y, join, z is what we still have there. So we get that one. So now looking at this one, we have an z on the second factor. We can factor it out using the distributive laws. This becomes x, meet, y, join, z, for which by assumption x, meet, y is the same thing as x, meet, z. And so we get x, meet, z, join, z by absorption. And again, there's some, there's some commutativity we'd also be using there as well. But by the absorption action, this gives us a z. And that then proves the second cancellation axiom. Well, the third one, I should say. All right, a few other properties. This one's a lot easier to do a few other properties of Boolean algebras I want to establish in this video before we end lecture 38. If x is an element inside of a Boolean algebra, then its complement is unique. There's an analogous statement for inverses like in groups. If you have an inverse, it's unique. We want that complements are unique as well. The proof is going to look a lot different though. So what we're going to do is suppose that we have two different complements for an element x. So let's say that x prime and not x are two different complements. Well, if x prime is a complement of x, that means that x join x prime is equal to zero and x meet x prime is equal to one. Now, if not x is a complement, that means that x join not x is equal to zero and x meet not x is equal to one. So that is we're assuming they're both complements. All right, so notice by assumption x join x prime is equal to zero, but x join not x is also equal to zero. So we get that those two, these two things in particular equal to each other. I'm going to write that to the side x join x prime is equal to x join not x, but also x meet x prime is equal to one. So we get this, but x meet not x is equal to one. So we get this one as well. So in particular, these elements are equal to each other. We get that x meet x prime is equal to x meet not x. So then by the cancellation law, the third one in particular, the one we just proved, we can cancel the x's and get that x prime equals not x. So two different complements for the same element actually have to be equal to each other. So complements are in fact equal to each other. Then an immediate consequence, mostly immediate consequence of this, is that if you take the double complement, you get back to the original elements. We had an analogous statement for that when it came to groups that the inverse of the inverse was the original element. And the proof of that is going to be the same proof here, that you use uniqueness. So if you walk like a compliment and you quack like a compliment, you have to be a compliment. So I want to show that x is a compliment for x prime, and therefore the double complement have to be x. So compare what happens here. So by assumption, x prime join x is equal to x join x prime, which is equal to zero, and x prime meet x is equal to x meet x prime, which equals to one. So x is acting like a compliment of x prime. And by uniqueness of compliments, that means x has to be a compliment of x prime so that we get that the compliment of the compliment was equal to the original element as well. And so as an immediate corollary of this, we get that zero is the compliment of one, and one is the compliment of zero. I'm not going to provide the proof of that one. I'm going to leave that as an exercise to the viewer. And so that's going to bring us to the end of lecture 38, which we've introduced distributive lattices, bounded lattices, and all of this leading to the idea of a Boolean algebra, for which this is the equivalent of like a field inside of order theory. The Boolean algebra is this most structured lattice we possibly have, much like a field is like the most structured ring that you could think of. We have all the axioms, associative, commutative, distributive. We have compliments in this situation. And our next and final lecture of this lecture series, lecture 39, we'll continue to develop the idea of a Boolean algebra. In particular, we're going to classify all finite Boolean algebras. Kind of sounds like what we did with finite fields. We could characterize all finite fields up to their order, and we can restrict their orders just to be powers of a prime. With Boolean algebras, we can do basically the same thing. All finite Boolean algebras will be isomorphic up to their order, and their order always has to be a power of two. I'll see you then. If you have any questions about any of these videos, please post your comments below. I'll be glad to answer them. Like these videos if you've learned anything. Subscribe to the channel to see more videos like this in the future. And if there's, since we are basically at the end of this lecture series, if there's more topics you want to learn about in the future, feel free to post those in the comments too. If there's some interest, I am glad to make additional videos on topics that are not part of a regular class.