 In this video, we're going to provide an example of an axiomatic system, and it's pretty simple and it's essentially artificial, meaning that I'm not trying to connect this to any broader mathematical theory. We're just going to take things that offer actually no context clues to what they are and we're going to proof stuff about them just using the logic. And so we don't want our, our proofs aren't going to be motivated by intuition. They're only going to be motivated by the axiomatic method. So let's get into that. With the axiomatic method, we have some undefined terms, some things that can be interpreted, but in themselves have no meaning. So these terms, we're going to call them fee, we're going to call them foe. What's a fee? What's a foe? It might make you think of giants and beanstalks, but really, these words are intentionally are chosen so that there's no context clues behind them. What's a fee? What's a foe? That could be anything, really. And I should mention that this example is derived from one in Wallace and West Roads to Geometry textbook. Now, fee and foe, they're things that will give some meaning to them in a moment, but there's a relation that belongs to fee and foe. And that's actually, that's a relation's name. So a fee can belong to a foe. So if you have a specific fee in a specific foe, there's some notion of belonging, a relationship. But what does it mean to belong? Now here I'm already starting to get some words here that try to give context, right? The word is trying to give us some interpretation, but be aware that by nature of the undefined terms, that's a cultural interpretation. That's not the logical interpretation there. There's none. It's just a relationship that we've called belonging. Now from these undefined terms, fee foe and belongs to, we then can come up with a new term. So a new definition. So if a fee belongs to a foe, then we say that the foe contains the fee. So containment is just the inverse relation to belonging. So we defined it using these undefined terms, right? So, and we could come up with more definitions we want to, but that's sufficient for this example. All right, so now we have undefined terms, we have our definitions. Again, we could add a lot more definitions we wanted to, but this is just a toy example here. Next, we're going to introduce for this axiomatic system, four axioms. These are going to be statements that are fundamentally true about our undefined and defined terms. They're true because I say they're true. And that's, that's all there is to it. Okay. And so we can have quibbles about whether they should be true or not. But for this logical system, they're true statements. And so everything that's true that will prove as a theorem is then derived from these axioms. So there, the truth of the theorems is then contingent upon the truth of the axioms, which as axioms were accepting to be true. So let me state what these axioms are with these four axioms. They're going to be statements about these fee foe and belongs to. So the first axiom says that there exist exactly three distinct fees in the system. We have three fees. Okay. The second axiom tells us that two distinct fees belong to exactly one foe. Okay. The third axiom is not all fees belong to the same foe. And then the fourth statement says that any two distinct fees contain at least one fee that belongs to both. And so when you read these things out of context, it can be very confusing why these axioms are accepted to be true. But again, this example as a toy example is intended to be that we don't have any context to why these things are true. We're just accepting that they are taken on faith that these axioms are in fact true. And for a fee foe logical system, it's always true because if these statements are true, then it's not fee foe theory. Okay. And so that's the third step of the axiomatic method. We have axioms. And so from these axioms, we can then start to prove theorems for which the first theorem you can see here on the bottom of the screen. And so while I usually will write the theorems more compact for this lecture series for these very first theorems, I do want to be very there's a lot of space here so we can focus on just one line at a time. All right. And for the most moment, I want to keep the axioms on the screen just so we can see what they are. All right. So the first theorem of fee foe theory says that each foe has exactly two fees that belong to it. Okay. So notice when we say here each each foe, this is a quantifier saying for all foes, there's exactly two fees that belong to it. So to prove this again, going back to our fundamentals of logic here fundamentals of mathematical proof writing, if we want to prove statement with a quantifier, what we're going to do is we're going to choose an arbitrary foe, we'll call it F for foe here. And so we want to prove that for F, there is exactly two fees that belong to it. Well, how do you get that? Well, axiom three, which we can still see on the screen right here, axiom three says that not all fees belong to the same foe. All right. So by axiom three, if we interpret that for the specific foe F, by axiom three, not all fees belong to F. Right. So there's some, there's some fee that doesn't belong to F. All right. That's almost seems like the opposite of what we're trying to do. We're trying to get, there's exactly two that belong to it. But notice here exactly two, exactly two says that there's at most two that belong to it. But it also says there's at most, there's at least two that belong to it. So that's what we're trying to work through here. Also axiom one gives us that there's exactly three fees in the system. So if we have our three fees, we'll call it A, B, and C. You know, these are our fees in the system. That's what axiom one gives us. And we have our foe F over here. We know that one of them doesn't belong to F. So we have to conclude that F does not have more than two fees belonging to it. That's what we wanted here. Right. You need at most two and at least two fees. We've now gotten them at most because as there's only three fees, if all three of them belong to F, then that would violate axiom, that would violate axiom three. Because axiom one gives us that there's three of them. All right. So why can't there be fewer than two? Now, suppose there were fewer than two fees that belong to F. This is going to be a approved by contradiction, right? If there are fewer than two, we're going to get a contradiction, which then thus implies there must be exactly two. Now, since there are exactly three fees, at least two fees do not belong to F. Let G be the unique foe for which these two fees have to belong to by axiom two. So using our little picture over here, if only one fee belonged to F, right? Or maybe no fees belong to F, right? That's a possibility right here. In particular, there's two fees that don't, at least two fees that don't belong to F. Now, by axiom two, which is no longer on the screen, let me just read it for you, any two distinct fees belong to exactly one foe. So these two fees, B and C, must belong to, there's one foe that they belong to, exactly one. And so we're going to call that one G for the moment if it doesn't belong to F. Well, then F and G are distinct since G contains fees that F doesn't. That makes them different, right? If they were the same foe, then they would be belonging to the same fees, which they're not. So we get that F and G have to be different for which then axiom four comes into play, which I'll remind you of that one. Axiom four says that any two distinct foes contain at least one fee that belongs to both of them. So there has to be one in common to them. So F and G, they contain a common fee, which must be the third fee that's not used to construct G. Because if we come back up to our model that we were drawing up here, right? We said that maybe A belongs to F, maybe it doesn't. We weren't certain, but the thing is G and F, since they're different, they have to contain something in common. And the only one who potentially belongs to F would be A. So it's going to have to be that A belongs to G as well. And this then leads to our contradiction, because then G was formed as the foe that contains B and C, but then we've now discovered that G also contains A. But axiom three says that no foe contains all of the fees. So we get a contradiction at this moment, which means that our assumption was wrong. The assumption was back here that suppose that F has fewer than two fees. So we've proven that F can't have three fees, but it can't have less than two. So it has to have exactly two fees, which then proves our very first theorem. This is the hardest of the theorems here. And the other ones are going to pull out very quickly from it. And again, this might seem strange, but just follow the logic here. There's nothing illogical about what we did. It might seem silly with these fees and foes, but from the axioms, we didn't get the statement that every foe contains exactly two fees. So let's look at another theorem. So a foe is completely characterized by the fees it contains. We kind of used a partial statement here in the previous proof. We had used the fact that if you're the same foe, then you contain the same fees. But could it be possible, again, we have our three fees, could it be possible that we have some foe that contains A and B, but then we have some other foe G that also contains A and B. So the same foe must have the same fees, which is to say that if you have different fees, you must be different foes. But what could it be that different foes could actually contain the same fees? It turns out that this theorem is going to tell us that actually no, that a foe is completely characterized by its fees. That is to say, if two foes are distinct, then they cannot contain the same fees. So to prove this, we're going to take two distinct foes. So F and G are foes, and they are not the same. To try to keep life a little bit easier, we're going to typically use capital letters to describe foes and lowercase letters to describe fees. These mnemonic devices help us keep track of what we're talking about. Now, as these two things are different, we'll get to that in a second. Now, both the foes contain exactly two fees. We get this from the previous statement. Now, the case we want to rule out is what if they contain the same two fees? What if F contains the fees A and B, but G does the same thing? They contain the exact two same fees. Well, this is going to contradict Axiom 2, which, remember, what is Axiom 2? Axiom 2 says that any two distinct fees belong to exactly one foe. So if A and B are two different fees, they only belong to one foe together. So this is a contradiction to Axiom 2 right here, since F and G would be two distinct foes containing the same the same pair. And therefore, F and G contain different fees. All right. And so we're basically using the contrapositive of Axiom 2 right here. So we know that foes contain exactly two fees. And now we know that different foes must have different fees. So we can actually identify a foe as a set of fees. So a foe is a set of fees. So because of the theory we've developed, it turns out we didn't have to have we didn't have to have foe be an undefined term. We allowed it to be, but in some essence we could define it to be a set of fees, but it has to satisfy certain conditions. But those conditions were given by the Axioms, right? The Axioms basically are the definitions of the undefined terms. And this relationship of belonging, if we think of foes as sets of fees, then it turns out belonging is just set containment with that interpretation. So we've been able to push the envelope pretty quickly with this theory. And so moving forward, we're going to use this sort of set theoretic version of foe to help us better understand this theory of fees and foes. We still have no idea what a fee should be, but at least we're like, okay, a foe is a pair of fees, okay? Two distinct foes contain exactly one fee. So to prove this, we're going to take two distinct foes, call them F and G. Now by Axiom 4, which Axiom 4 tells us that any two distinct foes contain at least one fee that belongs to both of them, all right? So F, it has two fees. We've seen this before. So we can identify F with just A and B. G will contain two fees, let's call them X and Y, but one of these fees has to be the same between F and G. Interesting. So it's kind of like, okay, what if we make an amendment here? Let's say that X belongs to both of them because Axiom 4 says there has to be at least one that's common to both of them. Now, if the intersection, maybe that's not the right word here, but maybe we'll just use it in terms of a set perspective, right? So if the intersection, because these are sets now, if the intersection is two, that actually would imply that F equals G, right? Because they would have to be the same elements. This would have to be X, Y. It would have to be G if their intersection was two, because they contain exactly two. But the fees that a foe contains uniquely to characterize that foe, we can identify foes with pairs of fees. This would contradict the previous theorem. So therefore, the intersection has to be one. That is to say, again, from a set theory, I think one of you distinct foes contain exactly one fee in common. And so then this very last, this will be our very last theorem for this lecture here, for this fee foe theory we've developed. So it turns out that there are exactly three distinct foes. We know by Axiom 1, there's exactly three distinct fees. But because of the theory we've developed about foes, as a foe is uniquely determined by its two fees, and the foes only overlap, well, they overlap exactly with one fee each. This is what we're going to get, okay? There are three fees. We can call them A, B, and C. These are our fees. Our foes are basically just pairs of, they're just pairs of fees. So one of the foes could be A and B. Another foe could be A and C. Notice that these foes do overlap at A. Another foe could be B and C, like so, for which the first and last overlap at B, the last two overlap at C. So that's good so far. And there's no other pair we could determine because with three fees, there are only three pairs of fees, right? This is just a simple combinatorics argument, right? Since the foes are always two fees, then we basically just get three choose two mini foes, which turns out to be three in this situation. And so there's exactly three foes, and these are what they have to be, basically. With whatever these fees are, and the belonging to relationship, the foes are basically just these pairs. And so with this, we've now essentially finished our fee foe theory. These four theorems we've done essentially has proven everything about these. Whatever interpretation we take for the fees, we then can interpret the foes and the belonging to be these sets of two fees each. And we've thus proven all of these. So what are the fees? Well, we'll talk about that in a future lecture. And particularly in lecture two, we'll give some interpretation to this fee foe theory, so it doesn't just sound like abstract nonsense. It really does have practical applications. But that does bring us to the end of lecture one. I appreciate you all for watching it. 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