a) x'z' + y'z'+ yz' + xy
b) (A'+B'+D')(A+B'+C')(A'+B+D')(B+C'+D')
I can't find a solution!
Assuming that you are using the notation in a boolean algebra it appears you could simplify this by grouping or expanding:
a) $x'z'+y'z'+yz'+xy = (x'+y'+y)z' +xy $
b) $(A'+B'+D')(A+B'+C')(A'+B+D')(B+C'+D')=(A'+B'+D')(A'+B+D')(A+B'+C')(B+C'+D')=(A'+D')(A+B'+C')(B+C'+D')$
I'll leave the rest for you to reason about.
Copyright © 2021 JogjaFile Inc.
Assuming that you are using the notation in a boolean algebra it appears you could simplify this by grouping or expanding:
a) $x'z'+y'z'+yz'+xy = (x'+y'+y)z' +xy $
b) $(A'+B'+D')(A+B'+C')(A'+B+D')(B+C'+D')=(A'+B'+D')(A'+B+D')(A+B'+C')(B+C'+D')=(A'+D')(A+B'+C')(B+C'+D')$
I'll leave the rest for you to reason about.