Simplification of: AB + A'C + BC in boolean algebra

Question

I am trying to understand the simplification of the boolean expression:

AB + A'C + BC

I know it simplifies to

A'C + BC

And I understand why, but I cannot figure out how to perform the simplification through the expression using the boolean algebra identities. I was wondering if someone could show me the steps needed to do this. Thank you in advance.

If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions. — Cameron Williams, Sep 05 '16 at 19:04
Thank you for the suggestion and I know how to use them I just still don't know how to come to the answer with equation simplification. — M. S., Sep 05 '16 at 22:12

score 5 · Answer 1 · answered Dec 02 '19 at 10:16

5

\begin{align} F & = AB + A'C + BC \\ & = AB + A'C + BC(A+A') \\ & = AB + A'C + ABC + A'BC \\ & = AB + ABC + A'C + A'BC \\ & = AB (1 + C) + A'C (1 + B) \\ & = AB + A'C \end{align}

answered Dec 02 '19 at 10:16

Aniket Kariya

151

score 4 · Answer 2 · edited Feb 12 '18 at 04:15

4

\begin{align*} &\mathrel{\phantom{=}}AB+A'C+BC\\ &=AB+A'C+BC(A+A') \quad \text{($A+A'=1$, Complementarity law)}\\ &=AB+A'C+ABC+A'BC\\ &=AB+ABC+A'C+ABC \quad \text{(Associative law)}\\ &=AB+A'C \quad \text{(Absorption law)} \end{align*}

edited Feb 12 '18 at 04:15

Ѕᴀᴀᴅ

34,263

answered Feb 12 '18 at 03:52

Anirban

41
1

1

The latest addition of the last but one line is $A'BC$ – tatojo Jul 06 '19 at 10:32

score 3 · Answer 3 · answered Sep 05 '16 at 19:02

3

The two expressions are not equal. The first expression is true when A and B is true and C false but the second is false in this case.

answered Sep 05 '16 at 19:02

laissez_faire

63

I might be mistaken but I believe that both equations return true in that instance. Because if A is true and C is false then A'C is true. Meaning the whole second equation is true because it is an OR. – M. S. Sep 05 '16 at 22:05
just for clarification because my notation isn't that common the two equations are (AB) + (A(¬C)) + (B*C) and the other one is (A(¬C)) + (BC) – M. S. Sep 05 '16 at 22:08
Okay, yes then I understand. I thought you had negation on the variable to the left of your prime character. One way to arrive at the simplified expression is: $AB+A(\neg C)+BC=AB(C+(\neg C))+A(\neg C)(B+(\neg B))+BC(A+(\neg A))=ABC+AB(\neg C)+AB(\neg C)+A(\neg B)(\neg C)+ABC+(\neg A)BC=ABC+AB(\neg C)+A(\neg B)(\neg C)+(\neg A)BC=BC(A+(\neg A))+A(\neg C)(B+(\neg B))=BC+A(\neg C)$ – laissez_faire Sep 06 '16 at 19:15
I know this is late, but thank you very much! – M. S. Nov 15 '16 at 02:59

score 0 · Answer 4 · answered Aug 03 '17 at 12:58

0

In this way , this can be simplified

                 LHS = AB+A'C+BC   
                 =  AB+A'C+BC (A+A')                  [ A+A'=1 ]
                  =  AB+A'C+ABC+A'BC
                  = AB+ABC+A'C+A'BC
                  =  AB (1+C)+A'C (1+B)
                  = AB+A'C                           [  1+C=1 ]
                                        =RHS..

answered Aug 03 '17 at 12:58

Priya yadav

11

Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts. – Daniele Tampieri Jan 22 '19 at 21:28

score 0 · Answer 5 · edited Jan 24 '20 at 18:00

0

$\begin{align*}&= AB+A′C+BC\\ &= AB+A′C+BC(A+A′)\\ &= AB+A′C+ABC+A′BC\\ &= AB+ABC+A′C+A′BC\\ &= AB(1+C)+A′C(1+B)\\ &= AB+A′C\end{align*}$

edited Jan 24 '20 at 18:00

answered Jan 24 '20 at 13:43

Ismail Khan

11

Gerard · Answer 6 · 2023-04-30T10:08:17.567

You can use a Karnaugh map (8 fields, no dividing lines this time).

See, for example:
https://nl.wikipedia.org/wiki/Karnaugh-diagram

In this diagram only one variable changes between two adjacent fields.

B B B' B 'Fields
A 1 1 0 0 ABC' ABC AB'C AB'C'
A' 0 1 1 0 A'BC' A'BC A'B'C A'B'C'
C' C C C'

You can see that BC is not adding new 1's'.

So: AB + A'C + BC = AB + A'C

score -1 · Answer 7 · edited Jul 08 '18 at 10:00

-1

a.b+a'.c+b.c
a.b+a'.c+b.c(a+a') {Complementary Law}
a.b+a'.c+(a.b.c+a'.b.c)
(a.b+a.b.c)+(a'.c+a'.c.b)
a.b(1+c)+a'.c(1+b) {As 1+c=1 and 1+b=1}
a.b+a'.c

edited Jul 08 '18 at 10:00

Martin Sleziak

53,687

answered Jul 08 '18 at 03:17

aritra das

11

score -1 · Answer 8 · edited Sep 23 '19 at 11:30

-1

=AB+A'C+BC =AB.A'C+BC(A+A') =AB+A'C+ABC+A'BC =AB+ABC+A'C+ABC =AB+A'C

edited Sep 23 '19 at 11:30

Especially Lime

40,828

answered Sep 23 '19 at 09:34

Caleb yego

11

Simplification of: AB + A'C + BC in boolean algebra

8 Answers8