 Welcome back to one more proof of a set identity using the algebra of sets method here. This is going to be a fairly lengthy and complex and involved proof and I just want to warn you about that because it's going to really require you to pay attention and put your thinking caps on and follow along very, very closely here. As we go through though, through this though, the one thing I do want to encourage you is that every step that you see here can be justified using one or more of these guys here, combined with these guys here and I will just point out that in the proof that's about to happen, this property shows up quite big here. That's one of the reasons we took pains to prove it using the choose an element method that A minus B is equal to A intersect B complement. So let's get to it and the purpose of this is to show you how complicated a set proof can possibly get but also yet at each individual stage at the micro stage is actually a fairly simple process. We just have to be a little creative and we have to learn how to know when to solve problems in a slightly out of the box sort of way. So anyhow, here we go. The proposition says that for any sets A, B and C that are subsets of a universal set, we have A minus B minus C is A minus C minus B minus C. You might not have expected that to happen. You might have thought that maybe I could regroup and this C showing up twice seems a little weird here. This just goes to show that set operations like minus don't mean the same thing as arithmetic minus that we use on numbers. So we're going to show this proposition here. We could do this with choose an element, but we're also we're going to try it here with set properties as well. So let's go to it. And here we are, let me just make sure my color is correct. There we are. And so A minus B minus C. Now, first of all, what is that equal to? Let's use a set identities that we know. So I'm going to go in here inside the first parenthesis and rewrite that as a intersect B complement minus C. And that's because of that basic property, the one that we proved. Now, on the next step, I'm going to use that property again, because what I have here is one large set. Just group it off and call it x if you want to minus C. So I could rewrite that using the very same property as A intersect B complement intersect C complement. Okay, because here's the first thing in the subtraction. And here's the second thing. And that's what the property says. It doesn't really matter that the first term in the subtraction is more complicated. It just says whenever I have anything minus C, it's going to be anything intersect C complement. So now I see some things that I might want to move forward here, you might see that I have a couple of intersections in their group. So let me try regrouping intelligently. Let me make this a intersect B complement intersect C complement. Okay, that looks pretty good. So we could go with several different directions from here. I'm just going to point out a few of these that don't end up going there, they're correct steps, but they don't end up giving us much traction. One way one direction we could go here is to say that this is equal to A intersect B union C, the whole thing complement. And that's through De Morgan's laws. And then we could say from there that a this is equal to A minus B union C. And that seems like it's moving us in the right direction, except I don't really have a law. This is what I can do from this point forward. We don't have a distributive property that distributes the set difference over a union. So we're kind of stuck at this point, those two steps are correct, and I can justify them by pointing to the theorems. But they just don't really seem to get us in the right direction, they get us to a place where we don't have an identity to work with. And I would prefer not to have to go invent one. So what we're going to do here is tread on something that I've warned against, which is working backwards. Okay, now working backwards can be a bad idea if you're starting a proof by assuming the conclusion of working backwards. Here we're not going to do that, we're just thinking ahead and realizing that in the end what we want all this stuff to be equal to, all this stuff to be equal to is A minus C minus B minus C. We would like for this line here to eventually equal this line here. So what I'm going to do is jump straight to the end and work backwards steps. And I'm going to do one thing just to keep my memory fresher. Here's where we are before we start into the backwards mode here. So what I'm going to do now is start with the end, the thing that I want to show and work backwards from it. And if I can make this equal this, then I'm done. And so that's where we're going to proceed from here and change color again. Actually, I'm in green for the next slide here. So let's look at the ending and start working backwards. So here's where I want to end. And I can rewrite that as there's two subtractions. Actually, there's three subtractions if you look at it. There's one inside here, one inside here, and then one on the outside. Let me just kind of do them one by one. That first grouping symbol could be rewritten as A intersect C complement. And the second grouping symbol could be written as B intersect C complement. And I am subtracting them. Okay, so what I have here again is one set minus another. So I could rewrite that whole thing as A intersect C complement, intersect B intersect C complement, the whole thing complement. Okay, you have to think of these things as independent objects here. It's this thing minus this thing. And so I would take the first thing intersect the complement of the second thing, no matter how big or complicated it gets. Now, let's try to work from there. And again, all that stuff is by that one property that A minus B is equal to A intersect B complement. So let's see what else I can use here. I'm looking at the second grouping symbol here. And I have a complement of an intersection. So De Morgan's laws can kick in on that. I'm not going to do anything right now with this grouping set right here. But I am going to do something to the second group that I'm intersecting with. And this is the same thing as B complement union C. There's actually a couple of steps I did there. One was to use De Morgan's laws to take the complement of an intersection to give me the union. And what I should have here is B complement and C complement complement to be technically correct there. But C complement complement is equal to C by another one of the laws one of the basic properties from theorem 5.20. So I have this so far. Now let's see what I can do with it. What I'm going to do now is think of this set here as one entire set. If you like, you can just rewrite it. Let's kind of group it all off and call it something like S if you want to. I have S intersected with a union. And so now I'm going to use the distributive property to distribute that set intersection over the union. And here we go. So that would be the S there is A intersect C complement. So this would be A intersect C complement. Intersect B complement. That whole thing union with A intersect C complement intersect C. Okay. And that is by the distributive property. Again, I'm thinking of, I'll go back and redo this. I'm thinking of this whole thing here as a single unit, a single set here. And so in the end, no matter how complicated this stuff inside gets, it's just a set intersected with a union. And that's what the distributive property allows me to do. Distribute to here, then distribute to here. Okay, now look inside each grouping symbol and I can rearrange some things. Let me switch back to green here. I'm going to just leave the first grouping set alone for right now. A intersect C complement intersect B complement. But I'm going to regroup using the associative property inside the second one here because I like what is lining up here. This is C complement intersect C. Now what is C complement intersect C? This is one of the basic properties of the empty set from theorem 5.18. And that says that this thing here, I'll just switch to red. This thing here is the empty set. C intersect is complement is the empty set. So let's rewrite what we can. And this will be A intersect C complement, parenthesis intersect B complement, union with A intersect the empty set. Now look again what I have. I have A intersect the empty set. Well that is this whole thing here A intersect C complement, parenthesis intersect B complement, union with the empty set. Now when I intersect with the empty set, I get the empty set. It's like multiplying by zero. When I union with the empty set, I get it's like adding zero. I get this entire line equal to A intersect C complement intersect B complement. Now let's just pause for a second. So that's kind of complicated. I told you it was going to be long and complex. It's fairly complicated but look where we are. I have, I'm down to here, A intersect C complement intersect B complement. Now page back, what we wanted to show and where we were was almost that exact same thing. In fact that's exactly where we are. I just have to do one extra thing to get that red box to show up. And that is to simply regroup what's inside and switch orders around. So through a combination of the associative and the commutative property for intersections, I can rewrite this as A. Sorry I didn't mean to put a parenthesis there. A intersect B complement intersect C complement. And that is where I was on the previous page. So I will just refer back to the previous page here. And so that actually, if you chain all this stuff together, this actually proves it. And I want to explain why because that may not be obvious at this point. Let me back up and just kind of read this from front to back here. So I started with A minus B parenthesis minus C. I got this through the, through the basic property that we proved. I got this through the basic property that we proved. And I got this through grouping through the associative property. Now this is this. Right now read backwards from here. This is equal to this. This is equal to this is equal to this is equal to this is equal to this is equal to this is equal to this which is finally equal to this. So although I was working backwards, I could very easily rewrite this to be completely going forwards and prove my result. And in fact, I just want to show you what that would look like if you rewrote this as a completely forwards proof. I would start with this statement here, the thing I'm trying to work with. And here is the combination of the first two steps that I used before using that basic property, regrouping commutative law to switch the ordering here, associative property to regroup these, these, these symbols here. And here's where if you're reading this from front to back, you might wonder what, what just happened? Where did this come from? Well, I can certainly this is equal to this here because of the basic properties of the empty set. Now I didn't come up with that by looking at this and thinking, hey, I should union all this with the empty set. I came up with that because I had been working backwards. And this is just kind of where I ended up. So I worked backwards, but I'm rewriting the proof going forwards. This is a very common problem solving trick as you work a little bit forwards and a little bit backwards and eventually you meet in the middle. This is the essence of the no show proofs way way back that we introduced in chapter one. So anyhow, just to see the rest of this proof, I have this, this empty set that is the same thing as this by the laws of the intersecting with the empty set continued. I had this is was the previous line. The empty set can be written as C intersect C complement. And now the rest of this is just using the properties that I have here. I'm just regrouping here. I'm distributing it's almost like refactoring more than distributing here. I'm using a law using De Morgan's laws and here is that basic property again. So in the end, I have this equals this. So that's a really long and complicated proof. And if you understood most or all of that, then you're to be congratulated. If not, then go back and go slowly over this and just realize that sometimes when you're proving identities, you can't just go forward, you go a little bit forward and a little bit backward, meet in the middle and then rewrite the whole thing as if it were going forward. And that's a perfectly valid problem solving technique because we're not assuming the conclusion holds. We're finding a connection between the given and the conclusion. And that's okay. Thanks a lot.