In boolean algebra, why is a+a'b = a+b?

Question

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 07:26
@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: https://courses.cs.washington.edu/courses/cse370/99sp/lectures/02-Comb/sld008.htm — C.J. Jackson, Sep 18 '16 at 18:05

score 21 · Answer 1 · answered Sep 18 '16 at 05:03

21

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

answered Sep 18 '16 at 05:03

JRN

6,566

Ben Grossmann · Accepted Answer · 2020-01-28T15:11:53.220

20

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

edited Jan 28 '20 at 15:11

answered Sep 18 '16 at 05:04

Ben Grossmann

225,327

Thanks. It's easy to forget that multiplication distribute over addition in boolean algebra. – C.J. Jackson Sep 18 '16 at 05:19
You mean the other way. – Leif Willerts Sep 18 '16 at 09:27
Huh, I knew the law but it hadn't sunk in for me that it's really addition distributing over multiplication. +1 for phrasing it that way! – user541686 Sep 18 '16 at 09:55
@Leif what is it that you think I mean? – Ben Grossmann Sep 18 '16 at 14:23
I meant OP means addition distributes over multiplication - that that's the hard to forget part. – Leif Willerts Sep 19 '16 at 19:51

In boolean algebra, why is a+a'b = a+b?

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