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In my Computer Logic class we were assigned the following problem:

Complete a truth table that has $3$ inputs $(A, B,C)$ and one output $(F)$. $F$ is asserted whenever $B$ or $C$ are asserted, but deasserted if both $B$ and $C$ are asserted. $F$ is asserted whenever $A$ is asserted and $C$ is deasserted. $F$ is deasserted in all other cases.

After which we are supposed to

Determine the sum-of-products equation for $F$. Show the terms instead of the minterm abbreviations (show $ABC$ instead of m7).

Finally, we are supposed to:

Prove that $F$ from question $2$ is equivalent to $F = BC' + B'C + AC'$ using proof by rewrite. HINT: remember the uniting law is the key to minimization.

Here is what I have:

1)

A    B    C    F
0    0    0    0
0    0    1    1
0    1    0    1
0    1    1    0
1    0    0    1
1    0    1    1
1    1    0    1
1    1    1    1

2) $F = A'B'C + A'BC' + AB'C' + AB'C + ABC'$

Number three is what is throwing me. I am using the axioms of boolean algebra, such as the Uniting Law, but so far all I have been able to do is the following:

3) A’B’C + A’BC’ + AB’C’ + AB’C + ABC’ = BC’ + B’C + AC’

A’B’C + A’BC’ + AB’C’ + AB’C + ABC’

A’B’C + AB’C + A’BC’ + AB’C’  + ABC’            Commutative Law

B’C + A’BC’ + AB’C’ + ABC’                      Uniting Law

A’BC’ + ABC’ + B’C + AB’C’                      Commutative Law

BC’ + B’C + AB’C’                               Uniting Law

That's as far as I am able to get. Did I screw up the truth table, resulting in incorrect minterms, or did I miss something?

Appreciate the help!

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  • $\begingroup$ I disagree with the close vote, here. Yes, it's an exercise problem and, yes, our usual response to those is "What did you try? Where did you get stuck?" and then close as "Unclear what you're asking." But allCrocs has explained exactly what they tried and exactly where they got stuck and asks specific questions about their solution attempt and where it was going. This is a reasonable question, not just a problem dump. $\endgroup$ – David Richerby Feb 6 '15 at 9:34
  • $\begingroup$ @DavidRicherby I appreciate that! $\endgroup$ – allCrocs Feb 6 '15 at 21:13
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The last line is wrong : 1 1 1 1 Should be : 1 1 1 0

But the equations following are correct.

For the simplification : BC’ + B’C + AB’C’ = BC' + B'.(C+AC') = BC'+B'.(C+A)=BC'+B'C+B'A

(I don't know the name of the rules used, there is some associativity, "X+X'Y = X+Y", etc...)

--- edit ---

Let's prove that X+X'Y = X+Y

Using :

  • A = A.B + A.B'
  • A = A+A
  • A + AB = A.(1+B) = A
  • A + A' = 1

X+X'Y = XY+XY'+X'Y = Y(X+X') + XY' = Y + XY'

Repeating (A=A+A) :

X+X'Y = X+X'Y + X+X'Y = X+X'Y + Y+XY'

= X + Y + X'Y + XY'

= X+XY' + Y+X'Y

= X + Y

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  • $\begingroup$ Thanks for the response! The last line was a typo, as shown in my minterms. Whoops. Thanks for the heads up. Now, I see what you have done, but I am still at a loss as to how to get from BC' + B'C + B'A to BC' + B'C + AC'. Any further tips or directions? $\endgroup$ – allCrocs Feb 6 '15 at 21:11
  • $\begingroup$ Oops, wrong term : BC'+B'C+AB'C' = C'(B+AB')+B'C = C'(B+A)+B'C = C'B+C'A+B'C ! $\endgroup$ – TEMLIB Feb 6 '15 at 23:34
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If you're in CDA 3103, I just got done with this. What I did when I got to where you were is take the C' out via the distributive law, then use the absorption law on the inside of the parentheses. Good luck!

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Rewriting TEMLIB's answer, since it should be C and not B:

BC’ + B’C + AB’C’

= BC' + C'(B+AB') --> Associativity Law

= BC'+ C'(B+A) --> Absorption Law

= BC' + B'C + AC' --> Distributive Law

And that should be your answer

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