 As we draw closer to the conclusion of our logic series, the final logical principle that we want to introduce in this lecture series is this idea of disproof. So far, we've been focused basically on all of our efforts on proving that statements say p is in fact a true statement. But what if p isn't true? What if actually p is false? Many times, when we're going on a statement, we don't actually know whether it's true or false. If it's true, we can provide a proof that it's true. But if the statement's false, we're never, ever, ever going to find a proof that shows it to be true. And so as we struggle to prove something, it might actually be the case that the statement was false the whole time. And so if the statement's false, then it's not true and we can't prove it to be true. But how do we prove something to be false? Well, it turns out if p is false, then its negation is actually true. And so all of our efforts to prove that something is true is you just switch to the negation here. We can prove a statement to be false by showing its negation is to be true. And so that's basically what disproof is all about. You just prove that the negation is actually true instead of the statement itself. Because one of those two things happens, either the statement p is true or its negation is true. And so if you have to prove or disprove, disprove is just a proof of the negation being true. And so let's review the different type of things that show up here. So like if you had an and statement, p and q here, well, if you wanted to disprove it, you would have to prove it's negation, so not p and q. But then the De Morgan laws come into play here, you get not p or not q. So if you wanted to disprove a conjunction, it turns into a disjunctive statement. And likewise, if you wanted to prove the negation of a disjunction and or statement by De Morgan laws, this becomes not p and not q. So to prove a conjunction or disjunction, we apply the principles we've learned already, but to disprove a conjunction or disjunction by the De Morgan laws, it comes down to proving a conjunction or disjunction, all right? So we don't really need to say much more about that, we've already considered it. What about an implication? If you have something like p implies q? Well, we've learned how you can use direct proof to prove it to be correct. If you negate it, then the way to think about this is you actually replace it with its logically equivalent form, not p or q because a implication is true, if its premise is false, it's vacuously true. And likewise, an implication is true, if its conclusion is true, it's trivially true. And so these two statements are logically equivalent, you can turn an implication into an or statement. And then by De Morgan laws, this becomes p and not q. So you have to prove a conjunction to negate a implication. Biconditional statements are similar because they're just an implication and another implication. So all of these connectives that we've had with primitive logical statements lead themselves to negation. We've discussed all these negation things in the past. In this video, what we're gonna focus on is the idea of a quantified statement that we want to disprove. So imagine we have the statement for all x inside of x, the property p of x holds, which remember this thing is essentially just a conditional statement because saying for all x inside of x implies p of x is really just as I suggested here, you take as your premise x belongs to x and then this implies the statement p of x. So how do you disprove this quantified statement here? Well, we're gonna think of it again as this quantified statement here for all x inside of x, p of x holds. Well, the rules for negating a quantified statement is the following, universal quantifiers turn into existential quantifiers and vice versa. So if you negate for all x inside of x, this is gonna become, there exists an x inside of x, and then since you're negating the p of x here, you get not p of x right here. So if you want to disprove a universal statement, you actually have to prove an existential statement where of course the property is negated in that situation. Well, how do you prove an existential statement? What you do is you have to provide an example because there exists just means there is at least one element that has this property. So we have to provide an object that doesn't satisfy the property. So one example is enough and since we're disproving, this is commonly referred to as a counter example. We provide one example where the property doesn't hold and that then negates the universal statement that was mentioned above. And so that's all that you have to do to disprove a universal statement. You provide a counter example. Now, this counter examples that you find are often associated to conjectures. Remember a conjecture is a statement that we do not know if it's true or false, although generally we believe the conjecture to be true, right? But we don't actually know because a proof hasn't been provided to us yet. So consider the following famous conjecture. Let k be a natural number, then the number two to the two to the k plus one is a prime number, okay? This conjecture was first mentioned by Pierre de Fermat in 1650. So that was a quite long time ago. Fermat was convinced that numbers of this form are always going to be prime. And why is that? Well, if you look at the first couple of numbers in the sequence, it's not too hard to see why you would believe that. Like if you take k equals zero, then you plug in zero for k there, two to the zero is one, two to the one is two plus one is three. The first number in this sequence is three. If you take k equals one, that'll simplify to be five. If you take k equals two, you end up with 17. k equals three gives you 257 and k equals four gives you 65,537. For which that one's not less, that's probably not as obvious to the viewer here, but that is in fact a prime number. So the first five numbers in this sequence in fact are prime. And so for this reason, Fermat conjectured that all of the numbers in this sequence are in fact going to be prime. Nowadays, we refer to these as Fermat primes. But of course, back in the 1600s, we didn't have very good computers. We didn't have very good calculators nowadays. So it was reasonable to think that all of these numbers were in fact going to be prime, but it turns out they're not. And to prove that, we can show that there does exist a k for which this number is not prime. And in particular, if you take the very next one, k equals five, I can show you that this is not a prime number. Take the number two to the two to the five plus one. Two to the five is 32. Two to the 32nd power is gonna be 4,294,967,296. If you add one, well, you just get that number right there. And believe it or not, this number does in fact, it factors, it factors as 641 times 6,700,417. So it is a composite number. It is not a prime number. So this then shows that the conjecture is false because there does exist a counter example. That's all you have to do. You have to provide one example that shows it's false and that makes it a false statement. This is not always a prime number. In fact, these are the only five numbers that we know these are Fairmont primes, right? These are the only five known Fairmont primes in existence. If you try, we just did n equals five. If you try n equals six, seven, eight, nine, 10, et cetera. Every explored number here, it turns out other than these five turns out to be composite. It's not proven in general, but actually the conjecture, the conjecture has flipped on its head. Modern day mathematics believes that these are the only five Fairmont primes in existence that every other number of this form is composite. But that has yet to been proven as well. It turns out there could always be a new Fairmont prime in the future that is yet discovered. That's the interesting thing about conjectures. We don't know. As counter examples are found, we often change the conjecture to reflect those counter examples, but we don't know. But this does in fact show that the above statement is false because it has a counter example. Let's do one involving sets here. So imagine we have the sets A, B, and C. We conjecture that if you take the set A minus B intersect C, this is equal to A minus B intersect A minus C. This equation right here seems reasonable. We've proven some properties of sets that are similar to this. This could be true. Now, if you weren't sure if it was true or false, what you would do is you would then try to prove it. If you think it's true, you're gonna try to prove it. Proving that two sets are equal to each other means you're gonna take one element inside of one and argue it's inside the other and then do that in both directions. So if I was gonna prove this to be true, this is what I would do. Now, just so you're aware of the proof that we're writing right now is incomplete. And in fact, this is a false statement. We'll provide a counter example in a moment. But if we tried to prove it, this is what would happen. I would take some element X that belongs to the set on the left. So let's say that X is inside of A, take away B minus C, of B intersect C. So what does it mean to be inside here? It means that X is inside of A and that X is not inside B intersect C. Well, by the De Morgan laws, if X is not inside the intersection of B and C, that means that X is not in B or, that or is an important word here, or X is not inside of C. Well, as there's really no significant difference between B and C here without the loss of generality, let's suppose that X is not inside of B. Well, since X is in A and X is not in B, that gets us that X is inside of A intersect B, all right? Notice our target though is we're trying to get that we're in A intersect B and A intersect C. Sorry, A minus B intersect B minus C. We've gotten that X is inside of A minus B, but how in the world are we gonna show that X is also inside of A minus C? Well, we know X is in A, so we need that X is not in C for that to work. But remember, we had this or statement before. We could make the assumption that X was not in B, but because of the or, we can't also get the assumption that X is not in C. So we're kind of hitting a roadblock right here. There's an issue in play that we got one of the sides, but we didn't get the other sides. And this is something that happens when you try to prove a statement. When you're struggling to write a proof, it means one of two things. Either we have a misunderstanding or we have some type of knowledge gap, some gap of understanding on what's going on here. So maybe it's just a hard proof and we haven't finished it yet. The other possibility is that the statement's actually false, which is why we can't prove it. And in fact, there's a counter example that exists and we're basically adjacent to it that our proof is now obstructed because we can't get past this logical step because there's actually a, well, there's a counter example existing right here. What if there exists some element in this situation where X is in, is not in B, but X is in C, right? Which case I'll never be able to prove this one even though I have this one right here. And so this gives me a hint on how I could construct maybe a counter example. So I wanna find, I have an element that's in A minus B, but then I wanna find an element that is in C, which means that I can grab an element that's in A and in C, but not in B. Scrolling down. Consider the sets A, which will contain one, two, three. B, which will contain four, five, six. And C, which contains one, three, five. Notice there is an element one, which lives in A and C, but never in B. It's exactly what I'm looking for. I bet you these sets will provide a counter example to the statement we had before. So note, if I take A minus B intersect C, well B intersect C is gonna be just five. That's the only thing that lives to both of them. And if you take five away from A, you took away nothing, you get back A. So A minus B intersect C is just the set one, two, three. On the other hand, if you take A minus B intersect A minus C, well A minus B is going to just be one, two, three because there's nothing in B to take out of A. On the other hand, A minus C does have something. C, it contains one and three, which you take away and you're left with just a two right here. One, two, three intersect two gives you back two. And so notice here that the set one, two, three is not the same thing as the set two. They disagree and therefore these two sets disagree as well. We then have found a counter example which shows that these two sets are not equal to each other in general. I mean, sometimes they can be, but that's the thing is sometimes they can maybe, but sometimes they can't, right? There could be examples of sets A, B, C where this equality holds, but we know in general it doesn't happen.