# In boolean algebra, why is a+a'b = a+b? [duplicate]

Looking at the truth table, they're equivalent:

A   B      (A+B)  (A+A'B)
-------------------------------
1   1       1       1
1   0       1       1
0   1       1       1
0   0       0       0


But what manipulation can one do using basic identifies and laws to show that they're the same?

• By A' you mean not A? I personally never used nor seen used that notation. I think $\overline{A}$ and $\lnot A$ are much more used around the world (or at least in the material I have seen). – Bakuriu Sep 18 '16 at 7:26
• @Bakuriu it's a notation that's used more in computer science, out of the necessity of not being able to do overbar in ASCII. See: courses.cs.washington.edu/courses/cse370/99sp/lectures/02-Comb/… – C.J. Jackson Sep 18 '16 at 18:05

Note with the laws of Boolean algebra, "addition" distributes over "multiplication" (just as multiplication would normally distribute over addition). Thus, we have $$a + (a'\cdot b) = (a+a')\cdot (a+b) = 1(a+b) = a+b$$
$A+A'B=A(1+B)+A'B=A+AB+A'B=A+(A+A')B=A+B$