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.