Boolean Algebra : Prove that [closed]

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.

Jerz

Why not just use a truth table to prove it? I think it is an identity. – Piffle

This question appears to be off-topic because it is about math, not programming. – Seneca

Answer 1

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

An

How did you manage to get the first expression? i.e. A+A'B = (A + A')(A+B) is there any method for that? – Jerz

@An we are getting extra term AB after you expand (A + A')(A+B) – Northampton

@Raptor How did you expand (A+A')(A+B)? – An

Answer 2

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

Song

which rule did you apply for every step? – Pastoralize

How did you right this: A.1 + A'B = A.(1+B)+A'B? – Handicraftsman

This is what I was looking for. It should be the accepted answer. – Begot

Answer 3

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

Northampton

I havnt studied de-morgan yet its the next topic :D thanks though i'll comeback after im done with it – Jerz

Its easy to understand. Just google it if u are too curious about it :) – Northampton