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I need help to resolve this problem, i have the following boolean function:

[(A.!C)+!(A.!C)].!(A.!B)

The Truth table is:

enter image description here

(please see this LINK TO wolframalpha for more detail)

Then the Sum-of-Minterms Form is:

A.B.C + A.B.!C + !A.B.C + !A.B.!C + !A.!B.C + !A.!B.!C

And the Product-of-Maxterms Form is:

(A + !B + C).(A + !B + !C)

¿Is this correct? ¿How do I demonstrate this? Because I tried to find the maxterms starting from the initial Boolean function, for example:

 [(A.!C)+!(A.!C)].!(A.!B)
= [(A.!C)+!(A.!C)].!(A.!B)
= (A.!C + !A + C).(!A + B)
= (!A + !C + C).(!A + B)
= (!A + 1).(!A + B)
= !A + B
= .....
= (A + !B + C).(A + !B + !C)  ???????????????

I do not understand how to find (A + !B + C).(A + !B + !C).

Please appreciate any help. Thanks.

Gaston Flores
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1 Answers1

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Let's call your expression $f$, $$ \DeclareMathOperator{\and}{~And~} \DeclareMathOperator{\or}{~Or~} \DeclareMathOperator{\nt}{~Not~} \DeclareMathOperator{\T}{{\color{blue}T}} \DeclareMathOperator{\F}{{\color{red}F}} f = \bigg((A \and \nt C) \or \nt (A \and \nt C)\bigg) \and \nt (A \and \nt B) $$

$$\begin{array} {c|ccc|c} & A & B & C & f \\ \hline \text{Row 1} & \T & \T & \T & \T \\ \text{Row 2} & \T & \T & \F & \T \\ \text{Row 3} & \T & \F & \T & \F \\ \text{Row 4} & \T & \F & \F & \F \\ \text{Row 5} & \F & \T & \T & \T \\ \text{Row 6} & \F & \T & \F & \T \\ \text{Row 7} & \F & \F & \T & \T \\ \text{Row 8} & \F & \F & \F & \T \\ \end{array}$$

Notice how your sum of minterms has 6 terms. Your truth table also has 6 "true" entries. To construct minterms, you just list the "true" cases:

$$\begin{align}f &= (\text{Row 1}) \or (\text{Row 2}) \or (\text{Row 5}) \or (\text{Row 6}) \or (\text{Row 7}) \or (\text{Row 8})\\ &= ABC \or AB\overline C \or \overline ABC \or \overline AB\overline C \or \overline A\overline BC \or \overline A\overline B\overline C\end{align}$$

To construct the product of maxterms, look at the value of $\nt f$. There are 2 cases when $f$ is false:

$$\begin{align}\nt f &= (\text{Row 3}) \or (\text{Row 4}) \\ &= A\overline B C \or A \overline B \overline C\end{align}$$

Now apply DeMorgan's: $$\begin{align} \nt\nt f &= \nt(A\overline B C \or A \overline B \overline C) \\ f &= (\overline A\or B\or \overline C) \and (\overline A \or B \or C) \end{align}$$

Your given answer for "product of maxterms" is the value of $f(\nt A, \nt B, \nt C)$. Perhaps you are working with a different definition.

DanielV
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  • Wow! great answer, thanks, but i do not understand something, so the correct form of product of maxterms is: (!A OR B OR !C) AND (!A OR B OR C) but this is not equal to A!BC OR !(ABC). – Gaston Flores Oct 03 '14 at 04:52
  • Please see this link, I construct the minterms from initial function using boolean rules (Truth table unused) – Gaston Flores Oct 03 '14 at 05:03
  • Let's call the answer you got $g$, so $g = (A + !B + C).(A + !B + !C)$. Check the value of $f$ and $g$ when $A = \text{true}, B = \text{false}, C = \text{false}$. You will see that they are different. $$~$$ As far as deriving the minterms algebraically, first find a sum-of-products. Then for each term $x$ that is missing a variable $y$ apply the rule $x = xy + x\overline y$. For example, you have derived $!A + B$. Since $!A$ is missing $C$, apply $!A = !AC + !A!C$. Keep doing that until all terms have all variables. – DanielV Oct 03 '14 at 05:19