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.

Answer 1

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

Answer 2

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

Answer 3

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