Simplifying boolean expression: $!(x!z+y!z+xy+z)$

Question

This is the expression:

', ! not
+ or

$((x'y'+z)'+z+xy+wz)'$ After some steps I can get $!(x!z+y!z+xy+z)$

How can I continue from here?

Answer 1

Using $x'$ may be a better idea as $x!z$ is confusing, do you mean "not $x$ and $z$," or "$x$ and not $z$"? Having said that, we have \begin{align} ((x'y'+z)'+z+xy+wz)' & = (z'(x+y)+z+xy+wz)' \\ & = (z'x+z'y+z+xy+wz)' \\ & = (z'x+z'y+xy+z)' \\ & = (z+x')(z+y')(x'+y')z' \\ & = z'(z+x')(z+y')(x'+y') \\ & = (z'z+z'x')(z+y')(x'+y') \\ & = z'x'(z+y')(x'+y') \\ & = (z'x'z+z'x'y')(x'+y') \\ & = z'x'y'(x'+y') \\ & = z'x'y'x'+z'x'y'y' \\ & = z'x'y'+z'x'y' \\ & = z'x'y'. \end{align} Now the third line is what you got as well. After that you apply de Morgan's once again, and simplify the remaining expression. Please let me know if any of the steps is not clear.

Answer 2

Start by using De Morgan's Law to get

$$ (x'+z)(y'+z)(x'+y')z' $$ Now, by distributing the $z'$ over the first term, you get $$ (x'z'+zz')(y'+z)(x'+y') $$ But $zz'=0$ (false), so we have $$ x'z'(y'+z)(x'+y') $$ Distributing $z'$ over the first remaining bracketed term, we have

$$ x'(y'z'+zz')(x'+y') $$ Again, $zz'=0$, so we have

$$ x'y'z'(x'+y') $$ In order for this to be true, we must have $x'=y'=z'=1$ (true), and if $x'=y'=1$, then $x'+y'=1$ too. Therefore, this reduces to $$ x'y'z' = (x+y+z)' $$