a do B is equal to B do a this is called commutative
commutative
property next associative property A + B
+ C is equal to Simply you can
interchange the bracket because all have
addition here so interchanging the
bracket nothing will change so a do B do
c can be written as a do B do c okay
next we have distributive property a +
BC can be written as a +
b into a + C so this can be split up
like this similarly a do B + C is equal
to you see the difference between these
two here first plus sign is coming
second we have a DOT product here first
dot is coming and second we have a plus
sign so here you have to split it as a
c so this come like this a do
B+ a do
c next we will see the absorption law so
here a + a b equal 2 what is common here
a so let us take a outside so it becomes
a into 1 + B what is 1 + B so remember
the Boolean law which we have studied
earlier what is a + 1 or 1 + a it is
equal to 1 that is if a is 0 or 1 any
value a can be but 1 + something will be
always = to 1 so in this case 1 + B is
here it means that it is equal to 1 so
1 so a + a b = to
a next we have a into a + b so this dot
product and plus sign has been changed
so a DOT A + B is equal to we can write
a into a + a into B so what is a do a
nothing but it is a because here it is a
multiplication symbol so a plus a b now
what is common here A A into 1 + B so
already we have seen 1 + B is equal to 1
so this is equal to a so you should
remember these two formulas also a plus
a b you can directly write it as a or a
a next we have a + a bar B so in earlier
we saw a + a b now it is in complemented
form a bar is here so we can write it as
a + a bar into a + b
so what is a + a bar that is 1 + 0 is 1
always 1 so 1 into a + b so that is
b so again we have a complemented a here
so we are simplifying it a do a bar + a
do B so what is a do a bar it is always
equal to 0 plus a do B so that is equal to
a next we will see about Duality theorem
according to this theorem from every
Boolean relation another Boolean
relation can be derived suppose if you
have a and symbol you convert that into
a r symbol if you have R symbol convert
that into a and symbol and one you can
change to zero zeros to one and a to
complemented a so by doing this you will
get another relation which is almost
one the most important theorem in
digital electronics is the Demar theorem
so it is named after the mathematician de
de
Morgan so he gave two theorem which is
very useful in digital
circuits let us see the first theorem
that is complement of a product see here
it is a product a means multiplication
so it is a product the complement of a
product means bar symbol is there AB bar
equal to sum of the complement so sum of
here sum is there compliment ment is a
bar + B bar so a bar equal to a bar + B
Bar second one is complement of a sum
here you see you have a product here
plus sign is equal to same thing change
the sign Al here product of the
complement so these two theorems are
very very useful without this one you
circuits so let us verify the demorgans
theorem first law let us take it AB bar
equal to a bar + B Bar let us check
whether this statement is correct or not
so take a and b we know the four
combinations for the two input variables
0 0 0 1 1 0 1 1 so four combinations we
we have find a bar B bar so what is a
bar just if it is zero make it is one if
it is 0 1 1 means 0 0 okay so we found a
bar next find B bar for zero you have 1
1 means zero
zero
now a let us find first what is a
multiply these two so 0 multiply these
two again zero if one input is zero
multiplication will get 0o 1 into 1 is 1
so we got AB now we will find AB
bar so just inverse of this one so three
zeros so you'll get three ones here one
so it should be zero here
next we will find a bar + B bar so we
have already found a bar B Bar
separately add it 1 + 1 is 1 1 + 0 1 0 +
1 1 0 + 0 zero now you see both are
same so we have verifi the first
law next let us verify the second law
so already we found a bar B bar now we
have to find what is A + B so add these
two 0 0 0 + 1 is 1 1 + 0 1 1 1 is
1 so complement this one A + B Bar you
will get so here you will get one
remaining three are
zeros now you have to find this one a
bar Dot B bar so you have to multiply these
these
two 1 into 1 you will get one so
remaining cases three cases you will get
zero so we can see that both are equal
also so let us summarize what we have
seen commutative property is a + Bal to
B + a a do B is = to B do a associative
property is A + B + C is equal to a + b
+ C that is you can change the
bracket then we have distributive
property a + b c is equal to a + b into
a + C A do B + C is equal to a do B + a do
do
c the important formula is the demarin
theorem a b bar equal to a bar + B bar
and a + b the whole bar equal to a bar + B
bar if you like the
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