Simplify $ab' + ( a'+b )c$

Question

Boolean algebra How to solve it. I am stuck after first step which is $ab' +(ab')'c$. After that I couldn't get the exact answer.

The answer is $ab' + c.$

You can use again De Morgan's law from here (in the other 'direction' than you used it. — Marc Dinh, Aug 07 '20 at 07:57
This is a particular case of $x+x'y=x+y$. Follow Marc Dinh's clue to prove this. — amrsa, Aug 07 '20 at 08:04
Tell me if this is the right approach. Y = ab' + (a'+b)c Y = ab' + ( a'+b) c Y= (ab'+(ab')') (ab'+c). Y= (ab'+c) — Utkarsh Sharma, Aug 07 '20 at 08:16

score 1 · Answer 1 · answered Aug 07 '20 at 07:51

1

HINT

It is $k+k'c=k(c+c')+k'c,$ where $k=ab'.$ Can you finish it now?

answered Aug 07 '20 at 07:51

user376343

Yeah. I did. I got ab'c + ab'c' + (ab')c' . After that -- c ( ab' + (ab')') + ab'c' . Then -- c + ab'c' – Utkarsh Sharma Aug 07 '20 at 08:04

score 0 · Answer 2 · answered Aug 07 '20 at 15:42

0

$$ab'+(ab')'c = (ab'+(ab')')(ab'+c)=1(ab'+c)=ab'+c$$

answered Aug 07 '20 at 15:42

Bram28

2 Answers2