Boolean Algebra/ Logic simplification

Discussion in 'Homework Help' started by Rbelec, Oct 7, 2016.

  1. Rbelec

    Thread Starter New Member

    Sep 28, 2016
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    0
    Can anyone help me understand how to simplify these ?

    1) barA*B*C + A*C

    2) bar all (A*B*C + A*barB*C + A)

    3) bar all ( barA*B*C + A*C)

    4) bar all ( barA + A*barB*C + ABC )
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    4,796
    Pick one and show your best attempt at it. That will let us see where you are going right and where you are going wrong. It's hard to help you understand something until we see how you might be misunderstanding it.

    What is the difference between "bar" and "bar all"?

    You might adopt the pretty widely used convention that an apostrophe indicates logical negation of the factor immediately to its left. So, if I'm understanding your notation correctly, this would mean you have

    1) A'*B*C + A*C

    2) ( A*B*C + A*B'*C + A )'

    3) ( A'*B*C + A*C )'

    4) ( A' + A*B'*C + ABC )'
     
  3. Rbelec

    Thread Starter New Member

    Sep 28, 2016
    4
    0
    for the first one:
    A'*B*C + A*C

    I considered using theorem X + XY = X considering X = C and Y = B. But I don't think this is possible because C is part of another variable (A*C)

    Same deal for theorem X' + XY = X' + Y if X' = A' X = A Y = C
    would I be correct in saying that if 2 terms are being multiplied they must be considered as 1 variable?
    Im stuck at this point.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,747
    4,796
    The Theorem X + XY = X works because of the distributive properly of AND over OR, so you have

    X·(T + Y) = X·1 + X·Y = X + XY

    Then we have the dominance property of T over OR which is that a logical T (True) OR'ed with anything is a logical T. So

    X·(T + Y) = X·T

    And finally we have the identity property of T with respect to AND which is that anything AND'ed with T is unchanged.

    X·T = X

    So combining all of this together, we have

    X = X + X·Y

    Notice that we can't follow this chain of reasoning with

    A'*B*C + A*C

    We can use the distributive property to write this as

    A'·B·C + A*C = ( A'·B + A )·C

    Now we must consider what we can (and can't) do with

    A'·B + A

    Let's tackle this one informally first. If A = T, then it doesn't matter what B is. If A = F, then the output is simply B. First and foremost, this means that if we can simplify this, that if has to involve both A and B somehow. Now consider the case when B = T. In this case, while not obvious, the output will be T regardless of what A is. Work at understanding why this is the case and then see if that will help you figure out how to manipulate things to show that this expression is equivalent to simply A + B.
     
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