Can someone help me with simplifying this Boolean term
$$(y + z')(x' + w) + (x + y')(w' + z)$$
$$x'= \mathrm{not} \quad x$$
I know it can be simplified into $1$, but I don't know how.
Can you please show me step by step and the laws which you have applied in simplifying?
Thank you very much!
Leave the second term as it is.
First,note that $(y+z')(x'+w) = (y'z)'(xw')' = (y'z+xw')'$. by De Morgan.
Also, by the law that $(a+b)(a+c) = a+bc$ , we get $((y'z + xw'))' = ((y'z+x)(y'z+w'))' = ((y'+x)(z+x)(y'+w')(z+w'))'$,
which then becomes again using De Morgan, $((y'+x)(w'+z))' + ((z+x)(w'+y'))'$.
Hence, we get in total: $$ (x+y')(w'+z) + ((y'+x)(w'+z))' + ((z+x)(w'+y'))' $$
By the rule that $a+a' = 1$ and $1+b=1$, it's easy to see that the above expression is $1$.
Main rules used : De Morgan, and $(a+b)(a+c) = a+bc$