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.