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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.

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  • $\begingroup$ If you get stuck in the future, try Karnaugh maps. They can help to simplify complicated boolean expressions. $\endgroup$ – Cameron Williams Sep 5 '16 at 19:04
  • $\begingroup$ 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. $\endgroup$ – M. S. Sep 5 '16 at 22:12
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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.

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  • $\begingroup$ 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. $\endgroup$ – M. S. Sep 5 '16 at 22:05
  • $\begingroup$ just for clarification because my notation isn't that common the two equations are (AB) + (A*(¬C)) + (BC) and the other one is (A*(¬C)) + (B*C) $\endgroup$ – M. S. Sep 5 '16 at 22:08
  • $\begingroup$ 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: $AB+A(\neg C)+BC=AB(C+(\neg C))+A(\neg C)(B+(\neg B))+BC(A+(\neg A))=ABC+AB(\neg C)+AB(\neg C)+A(\neg B)(\neg C)+ABC+(\neg A)BC=ABC+AB(\neg C)+A(\neg B)(\neg C)+(\neg A)BC=BC(A+(\neg A))+A(\neg C)(B+(\neg B))=BC+A(\neg C)$ $\endgroup$ – laissez_faire Sep 6 '16 at 19:15
  • $\begingroup$ I know this is late, but thank you very much! $\endgroup$ – M. S. Nov 15 '16 at 2:59
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\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*}

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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..
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  • $\begingroup$ Please use Math Jax in order to produce beautiful and beautifully formatted mathematical texts. $\endgroup$ – Daniele Tampieri Jan 22 at 21:28
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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

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