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