Is Demorgan's Theorem Applicable to This Boolean Expression?

[needhelp83]

$$\overline{ \overline{A}+ \overline{B} + \overline{A}B }$$

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

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

$$AB(A + \overline{B})$$

$$AAB + AB\overline{B}$$

ANSWER=AB

Just wanted to check. I haven't done this in a while

 

[wildman]

The third step is wrong, but you got the right answer.

 

[needhelp83]

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

 

[needhelp83]

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

 

[needhelp83]

Oops the equation should go as follows:

$$(\overline{A1}*\overline{A0}*\overline{B1}*\overline{B 0}) + \overline{A1}A0\overline{B1}B0 + A1\overline{A0}B1\overline{B0} + A1A0B1B0$$

 

[needhelp83]

Any suggestions?