 So let's do this. Let's do our first proof. Now I want to prove that a1 minus a2 equals a1's intersection with the complement of a2. How do we go about doing these proofs? I'm going to let you in on a little secret. All you have to do with proofs is to start by the definitions. Remember, the beams on this planet, we've decided that these are our definitions. Remember the definition of the subtraction of one set from the other. We define that as all the set of all elements x. Now those, remember, that's just a placeholder that is one of the elements inside of that. Such that x is an element of a1, but x is not an element of a2. And we define intersection as all the elements x such that x is an element of a1 and x must be an element of a2. I want to show you two more. If I take a2, it's complement. That's all the elements such that those elements are not an element of a2. And a2 is all the elements x such that x is not an element of a2's complement. We've decided that's how we define it. And that's all that because we have those definitions. We start with a few things that we all agree on. Those are our definitions. And all you can use in your proof is that once we've proven this, this goes into one of these and we can pull it out and we can use it in a proof. So we start with this empty vessel. We decide on a few definitions and we use those to prove the next thing. That goes into our little box and now we have a bigger box or more in the box. And we can use that in our next proof, in the next proof, in the next proof. So if you're struggling with how to construct a proof, you have to have this in your box. And I think a good thing to do is to have a huge board or then pieces of paper and you start writing. Number one, this is the first thing I ever learned that goes in my box. Is it something that we've all accepted as a definition? We take that, that goes in the box. You start filling up the box. You do your first proof. All you have to go on is what's inside your box. And the clues are in the box. You are going to find it there. Then, once you've proven something that goes in the box, put another tick with a little mark in it or whatever to show that there was a proof from the definitions. But that's now in the box. You're going to get another problem. You're going to be asked to do another proof. Go to your box. The clues are in the box and you fill that box up with more and more and more as time goes by. And eventually, proofs won't be that bad. So let's use this one. Now, if you have to prove an equality like this, it is in logic we have this sign that it implies in both directions. So I've got to prove this, this implies that and that implies this. Once I've done that, I have the proof. I've got to go in both directions. Now, let's go in this direction. Now, this is just a little thought process that's going to go on and then we'll do the proper proof. So how did we define a sub 1 minus a sub 2? Well, we define that as a set of all x such that x is an element of a 1. But and I should say x is not an element of a 2. That's how we defined it. But look at this, if x is not an element of a 2, that is how we defined that. So if it's not an element of a 2, it must be an element of its complement. That's all we're going to do. So this we can rewrite as x such that x is an element of a 1 and x must be an element of a 2 complement. That's why we defined reversal sets. Everything, all of those things have gone in my box and I could pull them out my box. And if I just look at this and I don't look at what is there, if I just see this as b and c or d and e or two sets, this is nothing but that definition right there. That definition right there. So I can say the following. A sub 1 minus a sub 2 is a subset of, that's all I've done. I have not proven more, is a subset of this, a sub 1 intersection, a sub 2 complement. As I say, don't get bogged down by what is written there. I might as well have written there as a say b's intersection of c is all of my elements x such that x is an element of b, but x is not n as an element of c. That is what is there, that is exactly what is there and we have this. Now let's go the other way around. Let's start with this definition. A, let's put this in a nice little box. We have that. Now we're going in the other direction. A sub 1 is its intersection with a sub 2's complement. How do we define that? That's all the elements x such that x is an element of a sub 1 and what is this? It must also be x is an element of a sub 2 complement. But if x is an element of a sub 2 complement, it cannot be an element of a sub 2. So this is exactly the same as x such that x is an element of a 1 and x is not an element of a 2. That's exactly the same thing. That is equality that we have there. If it's not an element and that's why remember we said we have the well-defined sets and we are going to assume that there is a universal set. That is how we construct what goes inside of our little box. We agreed that that is how we define sets. So those two are equal and if I look at this, this is nothing other than the definition of that. So I have that a sub 1 intersection a sub 2 prime is a subset of a sub 1 minus a sub 2. And I have this. Now we have something that I didn't think we've spoken about before is how do we define equality. And I'm going to use something other than a sub 1 and a sub 2. Let me do this. If I say that f, the set f is a subset of g and g is a subset of f that we define as f. I hope you can see the f equals g. We actually say that f equals g if and only if f is a subset of g and g is a subset of f then f is equal to g. Now I have this is a subset of that, but that's also a subset of that. Therefore, I have proven that this is true. So see how we used what we have known before. What was inside our box was all we had to prove. Now this wasn't yet inside your box. I should have told you this up here or before, but that is something that we all agree on so they can go inside the box. So there's something that we can go and that is our definition. These are our definitions. I can use all of those definitions in my proof. Now for my proof, I can say that by definition. Now there's an infinite number of ways to do this. I'm just going to show you one in your class. You might be asked to do it in a different way. Let something or let this be given that I'm going to write by definition. a sub 1 minus a sub 2 equals a set of all x such that x is an element of a1 and x. Now we use exactly that. It's not an element of a2 that by definition this is equal to that. We can write something like also by definition. I'll use just for definition. x is not an element of a2 equals x being an element of a2 complement. That was inside the box. That is how we decided because of our definition of a universal set we accept this. Therefore, I can rewrite this x an element of a1, x not an element of a2. Therefore this can be written as all the elements x such that x is an element of a1 and we can use that. x is an element of a2 complement. This is the definition of a sub 1 intersection with a sub 2 complement. We've decided that. Therefore, a1 minus a2 is contained within as a subset of a sub 1 intersection with a sub 2 complement. Nicely constructed. Conversely, we're going to start in the other direction. We're going to say a sub 1 intersection with a sub 2. This complement is the set of all x such that x is an element of a sub 1. We can use exactly that and x is an element of a sub 2 complement. Again, I'm going to say also by definition this, I'm going to do exactly what we did here. Eventually, I don't have space but you can see exactly what is this. Therefore, a sub 1 intersection with a sub 2 complement is contained within a sub 1 minus a sub 2. Then you can write since one set is contained within the other, because remember I can just do this. I see that as one set and that is one set. One is contained within the other and this one is contained within that one. I can write that out in nice neat words. Therefore, a sub 1 minus a sub 2 equals this set equals a sub 1's intersection with a sub 2 prime. Then there's various things you can do. At the end of the proof, you can put QED. You can put a little block or you can put two arrows pointing at each other. That shows the end of your proof. Let's see how neat we could just use what had gone before to use this. I can now ask you for a new kind of proof and now you can use this because you say by prior proof. If you have shown this, you can now use this inside of a new proof. Something else goes in your box and once that's in your box, you can use all of that for a new proof. The clues will always be what's inside your box. The problem is your box gets bigger and bigger and bigger and you have to remember what's inside that box. But that is where you pull out your proof.