Boolean Algebra : Prove that

Question

I was having trouble with the following problem in boolean algebra i.e.

A+A'B = A+B

I need to prove the above section. I mean its already reduced i can't reduce it further.

Why not just use a truth table to prove it? I think it is an identity. — Anirudh Ramanathan, Oct 21 '12 at 14:19
This question appears to be off-topic because it is about math, not programming. — Brad Larson, Mar 20 '14 at 18:14

score 13 · Answer 1 · answered Mar 20 '14 at 01:25

13

A+A'B = A.1 + A'B = A.(1+B)+A'B = A.1+A.B+A'B = A + B.(A+A') = A + B.1 = A + B

answered Mar 20 '14 at 01:25

rarief

131
1
3

which rule did you apply for every step? – Michele Dibenedetto Jan 24 '19 at 19:16
How did you right this: `A.1 + A'B = A.(1+B)+A'B`? – subtleseeker Apr 25 '19 at 18:31
2

This is what I was looking for. It should be the accepted answer. – Sachin Kumar Sep 17 '19 at 15:46

score 7 · Accepted Answer · answered Oct 21 '12 at 14:15

7

A + A'B = (A + A') (A + B) = 1 (A + B) = A + B

answered Oct 21 '12 at 14:15

timrau

22,578
4
51
64

How did you manage to get the first expression? i.e. A+A'B = (A + A')(A+B) is there any method for that? – kishanio Oct 21 '12 at 14:18
@timrau we are getting extra term AB after you expand (A + A')(A+B) – Afaq Oct 21 '12 at 14:21
@Raptor How did you _expand_ (A+A')(A+B)? – timrau Oct 21 '12 at 14:23

score 1 · Answer 3 · answered Oct 21 '12 at 14:28

1

First taking NOT on both sides and then apply De-Morgan's Law on both sides:

L.H.S=

(A+A'B)'
=(A'.(A'B)')
=(A'.(A+B')) //again applied de-morgan's law in previous step
=(A'.A + A'B')
=A'B'

also apply De-morgans on RHS
(A+B)'
=A'B'

Thus LHS = RHS

answered Oct 21 '12 at 14:28

Afaq

1,146
1
13
25

I havnt studied de-morgan yet its the next topic :D thanks though i'll comeback after im done with it – kishanio Oct 21 '12 at 14:35
Its easy to understand. Just google it if u are too curious about it :) – Afaq Oct 21 '12 at 14:36

Boolean Algebra : Prove that

3 Answers3