Simplifying Boolean Expressions with DeMorgan’s law

Question

I need help simplifying the following Boolean expressions using DeMorgan’s law:

a) [ (AB)' + (CD)' ]'

and

b) [(X+Y)' + (X+Y') ]'

Please show some steps so I can do the other ones myself

Solution 1

a) First step is the outermost negation: distribute it.

((AB)')'*((CD)')'

You see we have double negations which means the expression itself. (p')' = p therefore

ABCD

[ (AB)' + (CD)' ]'  --> ABCD

b)

Distribute the outermost negation:

((X+Y)')'(X+Y')'

get rid of the double negation:

(X+Y)(X+Y')'

again, distribute the negation (the one at the outer part of the expression):

(X+Y)(X'Y)

When you distribute (X+Y), we get

XX'Y + YX'Y

Since there is XX' in the first part of disjunction, the expression XX'Y equals to 0 (False). Multiple instances of the same thing in an expression is the same thing itself. ppp = p. Therefore: 0 + YX' --> YX'

[ (X+Y)' + (X+Y') ]' --> YX'

Im sorry for non-formal language:) hope it helps.

Solution 2

Steps are included:

a: [ (AB)' + (CD)' ]' = (AB)'' * (CD)'' = (AB) * (CD) = ABCD

b: [ (X+Y)' + (X+Y') ]' = (X+Y)'' * (X+Y')' = (X+Y) * (X'*Y) .. Simplifying this further relies on the distributive property.

Comments

[studiosfx]

Thank you. I tried to do this [(X+Y)' + (XY') ]' and I got X'Y' . Could you please double check if it's ok?

[studiosfx]

Thank you. Could you verify that [(X+Y)' + (XY') ]' = X'Y?

[Varaquilex]

I think the answer is right. Also tried it from an online De Morgan Calculator. You can check it as well, just type !(!(a+b) + (a+ !b)) for expression and you'll get a truth table. About your answer, could you double check the (X+Y')' negation again? Perhaps you might have missed the negation of Y.