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Can someone help me with this expression, and how to simplify that with all steps? Im kind of lost in those, and i have exam in 2 days.

Thank you

enter image description here

Expression

The result of expression is: F = B + C * A'

ItzMeN0n
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  • ...Or you have to simplify the expression for homework or a take-home exam due tomorrow, and have waited until the last minute...? – amWhy Nov 20 '16 at 18:34
  • @amWhy I dont have homeworks on my school so no. I just need help with this expressions, doesnt need to be that expression, i gave that expression because is on book, but in the book doesnt explain how we can solve that, it gives us only the result. – ItzMeN0n Nov 20 '16 at 18:51
  • You need to show some sort of context/effort. E.g. Surely you know that $A + \bar A = 1$. What other axioms or properties do you that can be used to tackle this, or another part, of the expression, e.g.? – amWhy Nov 20 '16 at 18:54
  • And if you know what the result is, you should put that in your question, too. – amWhy Nov 20 '16 at 18:55

1 Answers1

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I assume that you can use the usual axioms of Boolean algebras
(that these are bounded complemented distributive lattices).
Let me use $A'$ for your $\bar{A}$ (it's simpler in MathJax).

Now maybe you know, or are able to proof, that for for every $A,B$ in some Boolean algebra

  1. $A + AB = A$;
  2. using the above equality, you can also show that $A + A'B = A + B$.

And then, of course, as amWhy pointed in a comment, $A + A' = 1$ (this is also one of my original assumptions), and using several laws (starting with distributivity), you can show that $$(A+B')(A'+B) = AB + A'B'.$$

Now \begin{align} (A+A')B + ABC' + (A+ B')(A' + B)C &= B + ABC' +(AB + A'B')C\\ &= B + AB(C+C') + A'B'C\\ &= B + AB + A'B'C\\ &= B + A'B'C\\ &= B + A'C, \end{align} where in the last two equalities I used the results numbered 1. and 2. above.

amrsa
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