Welcome to the next video. In the Boolean series. In this video I'm going to show you what Boolean expressions are and how we can use the laws of logic to make them simpler. But I've shown you a lot of Lowe's and I've shown you a lot of operations, but what are actually Boolean expressions? But Boolean expressions are basically just input values. Peace, QS, and Rs combined together with all these logical operations, so different letters representing different input values and combined together with these logical operations just like in algebra for example. P&Q XRR is a Boolean expression, another somewhat more complicated Boolean expression is something like P&Q. Call Not Q&R Or P if then K. And not Q. If an only if R, so whichever operation you want to put together, and then how many inputs are there, it is again different for each expressions. As I mentioned, these different expressions will carry out different instructions for the computer, so the computer can do different things. And allow through certain input combinations and stop certain input combinations from going through in the circuit. Now let's talk a little bit about the rules of the brackets. So if I have got the expression of not P&Q, and if I have got the expression of not P&Q, what is the difference in here? What am I doing by placing the bracket? Well, just as in algebra by pressing a bracket somewhere, I'm emphasizing priority in this expression. The knot is only applied to pee and to be able to calculate the. Overall, output of this expression. I need to calculate not P and combine it together with the key using the end operation as opposed to hear hear the bracket is applied to the P&Q. So I need to calculate the P&Q 1st and then the note is applied to the all of it with the bracket. So once I found the P&Q output values then I need to invert them. I need to apply the not operation to those output values to get to the whole output so it will be very different in the two different cases. Similarly, if I have got P&Q. X or R or P&QX or R. The bracket tells me what I need to do first. Here I need to use the end operation and combined together P&Q then find the output and combined it together with the I using the axe or while in this case is the opposite way around. I need to use the exit gate combined Q&R together 1st and then use the end operation to combine output from here. Which P to get to the final output of the overall Boolean expression. Now the last thing I'd like to show you in this video is how we can use these laws of logic to reduce the Boolean expressions. So I have this expression not P or not Q. So how can I use the lose of logic to reduce this expression? Now I can use the Morgan low to distribute the not over the bracket. So what does the De Morgan do? I can break it up into not P and not not. Cute and then I can use the double negation and applied the not not key so that gives me not P&Q. Well, I think he would like to agree with me that instead of this bracketed expression, this expression is rather similar. One more example, not. Key or P? Or Not P&Q What I have here is key or P and I have here not P&Q. So I have got the same things in here but I have got them in the opposite order so why can do? First I can apply the commutative low and bring them up in the same kind of order. Then what I can use next is the Morgan loads to distribute the not inside the bracket. So that gives me not P and not Q. Or Not P&Q now what I have in here now. It's like in algebra you spot. That the first term here. Is the same, so you can do something called in algebra, factorization and in this case I can use the backwards operation. The backwards version of the distributive law so I can bring out the note P. And the end. And what remains is the not Q. Or key. Now what do I know about not Q or key? But I know about Nokia or cubes that that is always true because it doesn't matter which not Q or Q is force, the other will be always true and true. Or force always gives you true. So this is also the same as not P and true. And. Not P and true. That is always equal to not pee. I can apply the identity law here, which states that P and two is always P for the special case of not PN 2 is always not paying. So this long complicated expression is actually nothing else but not P. I hope that you now have a good idea of how to use the lose of logic to simplify Boolean expressions in the next slide you will have some questions to allow you to do some practice on your own and you will find the answers to these questions shortly after. So these are the practice questions. And here are the answers.