Is Demorgan's Theorem Applicable to This Boolean Expression?

  • Thread starter Thread starter needhelp83
  • Start date Start date
  • Tags Tags
    Theorem
AI Thread Summary
Demorgan's Theorem can be applied to simplify the Boolean expression provided, leading to the correct answer of AB. The initial steps included some errors, particularly in the third step, which was identified and corrected. A user expressed difficulty in simplifying a different equation using a K-map and inquired about the application of exclusive or. The revised expression was clarified, and users were encouraged to provide suggestions for further simplification. Overall, the discussion focused on the application of Boolean algebra principles and the importance of accuracy in simplification steps.
needhelp83
Messages
193
Reaction score
0
\overline{ \overline{A}+ \overline{B} + \overline{A}B }

\overline{ \overline{A}+ \overline{B} } * \overline{ \overline{A}B }

\overline{ \overline{A}}* \overline{\overline{B} } * (\overline{ \overline{A}}+\overline{\overline{B }})

AB(A + \overline{B})

AAB + AB\overline{B}

ANSWER=AB

Just wanted to check. I haven't done this in a while
 
Physics news on Phys.org
The third step is wrong, but you got the right answer.
 
I see the mistake. It should be:

\overline{ \overline{A}}* \overline{\overline{B}} * (\overline{ \overline{A}}+\overline{B})

AB(A + \overline{B})

AAB + AB\overline{B}

(A)B + A(0)

AB

Thanks for pointing that out
 
I put this equation in a K-map and I was unable to simplify it. Is there anyway to do this with exclusive or? Thanks for the help

\overline{A1}\overline{A0}\overline{B1}\overline{B0} + \overline{A1}A0\overline{B1}B0 +
A1\overline{A0}B1\overline{B0} + A1A0B1B0
 
Oops the equation should go as follows:

(\overline{A1}*\overline{A0}*\overline{B1}*\overline{B 0}) + \overline{A1}A0\overline{B1}B0 + A1\overline{A0}B1\overline{B0} + A1A0B1B0
 
Any suggestions?
 
Back
Top