De Morgan minimization

Question

De Morgan

$$y=\overline{a+\overline{b(\overline{c+d})}}+\bar{b}$$ $$y=\bar{a}(\overline{\overline{b(\overline{c+d})}})+\bar{b}$$ $$y=\bar{a}(b(\overline{c+d}))+\bar{b}$$ $$y=\bar{a}(b\overline{cd})+\bar{b}$$ $$y=\bar{a}b\overline{cd}+\bar{b}$$

This is as far as I've got on my simplification:

$$x+\bar{x}y=1+0\cdot1~or~0+1\cdot1$$

So it might be

$$1+0~or~0+1=1$$

But

$$\bar{a}b\overline{cd}+\bar{b}=\overline{acd}(b+\bar{b})=\overline{acd}\cdot 1=\overline{ac}d$$

I can't understand why it's will be \$y=\overline{acd}+b\$, so how I must to minimize this \$b\$?

Maybe I solved it! a'bc'd'+b'= (a+b'+c+d)'+b'= ((a+b'+c+d)b)'= (ab+b'b+cb+db)'= (ab+1+cb+db)'= ((a+c+d)b)'= (a+c+d)'+b'= =a'c'd'+b' Is it correct?

Answer 1

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.