boolean-algebra
I'm struggling with evaluating this one:
\$(\overline ABC\oplus A\overline B) + (\overline AB)\$
I've got to
\$A\oplus B + A\overline B \overline C +\overline ABC\$
But how do I prove that \$A\overline B \overline C +\overline ABC = 0\$
Edit: The answer is \$A\oplus B\$
You did it wrong by connecting the output of the first stage to the output of the second stage. The correct way would be :
TOTAL = 4 + 3 + 2 = 9 CMOS pairs
De Morgan $$y=\overline{a+\overline{b(\overline{c+d})}}+\bar{b}$$ $$y=\bar{a}(b(\overline{c+d}))+\bar{b}$$ You are correct up to this point. Based on the comments, you are correct for the rest, but you fooled up the mathjax syntax as you deMorganed. Corrected we have: $$y=\bar{a}b\bar c \bar d+\bar{b}$$
You are also headed in the correct direction: $$X+\bar{X} Y = X + Y$$
Which is the redundancy law. The inclusion of \$X\$ means the \$\bar X\$ in \$\bar X Y\$ is redundant.
Best Answer
So the Expanded equation is
Taking Common factors out
1+ anything is always 1
So we Have
A(bar)B+AB(bar)which isA(exor)B