Help needed with Boolean Algebra and DeMorgan

Thread Starter

randb

Joined Dec 29, 2012
16
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?
 

amilton542

Joined Nov 13, 2010
497
\( Y = [(AB')(A'C)]' \)

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

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

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 edited:

jjw

Joined Dec 24, 2013
551
\( Y = [(AB')(A'C)]' \)

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

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

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')
 

WBahn

Joined Mar 31, 2012
26,141
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?
 
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