0
$\begingroup$

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'

$\endgroup$
  • $\begingroup$ ...Or you have to simplify the expression for homework or a take-home exam due tomorrow, and have waited until the last minute...? $\endgroup$ – Namaste Nov 20 '16 at 18:34
  • $\begingroup$ @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. $\endgroup$ – ItzMeN0n Nov 20 '16 at 18:51
  • $\begingroup$ 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.? $\endgroup$ – Namaste Nov 20 '16 at 18:54
  • $\begingroup$ And if you know what the result is, you should put that in your question, too. $\endgroup$ – Namaste Nov 20 '16 at 18:55
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Well of course the first equality listed above is just one of the absortion laws... $\endgroup$ – amrsa Nov 22 '16 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.