Simplifying Boolean Expressions with DeMorgan’s law

boolean-expression demorgans-law
studiosfx picture studiosfx · Feb 2, 2013 · Viewed 12.5k times · Source

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

Answer

Varaquilex picture Varaquilex · Feb 2, 2013

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.

Related questions

Boolean Algebra Simplification of (x'y'+z)'+z+xy+wz

How would you simplify the following? I'm having a bit of trouble with the first part with negation. How would DeMorgan’s Theorem be applied here? (x'y'+z)'+z+xy+wz Please provide answer in detail. Update: The complete …

Read More
DeMorgan's law and C++

For each of the following write the equivalent C++ expressions, without any unary negation operators (!). (!= is still permitted) Use DeMorgan's law !( P && Q) = !P || !Q !( P || Q) = !P && !Q For !(x!=5 && x!=7) !(x<5 || …

Read More
Python `if x is not None` or `if not x is None`?

I've always thought of the if not x is None version to be more clear, but Google's style guide and PEP-8 both use if x is not None. Is there any minor performance difference (I'm assuming not), and is there …

Read More