Finding the Minimal Sum of a Logic Function

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I have the following logic function for which I was asked to find the minimal sum form.

$F = \bar{A}\bar{B}\bar{C} + \bar{A}B\bar{C} + AB\bar{C}$

Focusing on the second and third terms, I chose to pull out $B$ and then $\bar{C}$, allowing me to eliminate the $A$ terms:

$B(\bar{C}(\bar{A} + A))$

$B(\bar{C}(1))$

This leaves me with the following:

$F = \bar{A}\bar{B}\bar{C} + B\bar{C}$

I don't have a formal background in logic/algebra and so I'm a bit stumped as to what to do now. Assuming I did it right, a logic table for the original function, for which Wolfram Alpha confirms, the solution should be:

$F = \bar{A}\bar{B}\bar{C}$

I believe I'm only one transformation or two away from the answer, but am not sure what identity rule/trick I'm missing to solve this last step (or if I'm just outright wrong). I would appreciate any help with seeing the remaining steps or a potential lapse in my solution logic.

EDIT:


My claim the final equation above as the desired solution was wrong. I needed to factor out the $\bar{C}$ again and then distribute over the $\bar{B}$ for the inner term, which would yield the answer:

$F = \bar{A}B + \bar{A}\bar{C}$

Better yet, look at M.P.'s answer as it takes the smarter first step of factoring $\bar{C}$ out from all three terms in the initial equation, yielding a much more straightforward solution. Cheers!

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You can do one more step:

\begin{align} &&F ={}& \bar{A}\bar{B}\bar{C} + \bar{A}B\bar{C} + AB\bar{C} \\ \tag{Factorize $\bar{C}$.} \\ &\Leftrightarrow& F ={}& \left(\bar{A}\bar{B} + \bar{A}B + AB\right)\bar{C} \\ \tag{Factorize $B$.} \\ &\Leftrightarrow& F ={}& \left(\bar{A}\bar{B} + \left(\bar{A} + A\right)B\right)\bar{C} \\ \tag{$\bar{A} + A = 1$.} \\ &\Leftrightarrow& F ={}& \left(\bar{A}\bar{B} + B\right)\bar{C} \\ \tag{$\bar{A}\bar{B} + B = \bar{A} + B$.} \\ &\Leftrightarrow& F ={}& \left(\bar{A} + B\right)\bar{C} \\ \tag{Distribute $\bar{C}$.} \\ &\Leftrightarrow& F ={}& \bar{A}\bar{C} + B\bar{C} \text{.} \end{align}

To obtain $\bar{A}\bar{B} + B = \bar{A} + B$ just note that:

  • if $B = 1$ then $\bar{B} = 0$ and $\bar{A}\bar{B} + B = B$
  • if $B = 0$ then $\bar{B} = 1$ and $\bar{A}\bar{B} + B = \bar{A}$
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Some handy laws here are:

Adjacency

$PQ+PQ'=P$

Reduction

$P+P'Q=P+Q$

Generalized Reduction

$PR+P'QR=PR+QR$

Applied to your statement:

$$A'B'C'+A'BC'+ABC' = \text{ (Adjacency)}$$

$$A'C'+ABC' = \text{ (Generalized Reduction)}$$

$$A'C'+BC'$$