Demorgan problem

(( AB') (A'C))' One broke bar recd (AB')' + (A'C)'

then broke the bar again and recd A'+B+A+C' A'+a =1 1+ other variables =1?

Book states answer is (A+B') ( A'+C)

Where did I go wrong?

 

What's wrong with your title ?

 

Why are you applying boolean operators to expressions alone?

 

The answer in the book is wrong if the original expression is correct.

 

\( Y = [(AB')(A'C)]' \)

\( Y = (AB')' + (A'C)' \)

\( Y = (A'+B)(A+C') \)

Which is the same as you've got.

You're operating on expressions alone though, which is O.K, but could you type this "question" in it's entirety because something's missing.

 

\( Y = [(AB')(A'C)]' \)

\( Y = (AB')' + (A'C)' \)

\( Y = (A'+B)(A+C') \)

Which is the same as you've got.

You're operating on expressions alone though, which is O.K, but could you type this "question" in it's entirety because something's missing.

Last expression is missing + should be (A'+B)+(A+C')

 

Applying DeMorgan's theorem and Boolean algebra to the expression

results in _____.

 

Thank you for your responses and help

 

Last expression is missing + should be (A'+B)+(A+C')

Oops, yes you're right. I missed that one somehow.

In all fairness, I really don't know. Maybe someone else could help you.

 

Demorgan problem

(( AB') (A'C))' One broke bar recd (AB')' + (A'C)'

then broke the bar again and recd A'+B+A+C' A'+a =1 1+ other variables =1?

Book states answer is (A+B') ( A'+C)

Where did I go wrong?

Did you go wrong?

Let's check it out. You are claiming that the expression

((AB') (A'C))' = 1

This is of the form

(FG)' = 1

which is

FG = 0

F = AB'
G = A'C

If A is False, then F is False and FG is False.
If A' is False, the G is False and FG is False.

What does that tell you?