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I have some questions in a professor's material and in that questions there are the mathematical symbols for product and sum receiving numbers as parameters as following:

∑(0,3,4,6) =∏(1,2,5,7).

The question ask me to prove that for circuit with three inputs

∑(0,3,4,6) =∏(1,2,5,7).

This is not clear to me.

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    \$\begingroup\$ talk to your professor \$\endgroup\$ – jsotola Dec 5 '18 at 3:50
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\$∑\$ stands for SoP or Sum of Products. The 1's in a truth-table.

\$∏\$ stands for PoS or Product of Sums. The 0's in a truth table.

To illustrate the process: \$∑(0,1,4,7) = ∏(2,3,5,6)\$.

SoP (Minterms)

$$∑(0,1,4,7) = \overline A \overline B \overline C + \overline A \overline B C + A \overline B \overline C + ABC$$

Simplify:

$$∑(0,1,4,7) = \overline A \overline B + \overline B \overline C + ABC$$

PoS (Maxterms)

$$∏(2,3,5,6) = (A + \overline B + C)\ (A + \overline B + \overline C)\ (\overline A + B + \overline C)\ (\overline A + \overline B + C)$$

Multiply it out:

$$(AA + A \overline B + AC + A \overline B + \overline B \overline B + \overline B C + A \overline C + \overline B \overline C + \overline C C) \ (\overline A + B + \overline C)\ (\overline A + \overline B + C)$$

Simplify:

$$(A + \overline B)\ (\overline A + B + \overline C)\ (\overline A + \overline B + C)$$

Multiply it out:

$$(A \overline A + A B + A \overline C + \overline A \overline B + \overline B B + \overline B \overline C)\ (\overline A + \overline B + C)$$

Simplify:

$$(A B + A \overline C + \overline A \overline B + \overline B \overline C)\ (\overline A + \overline B + C)$$

Multiply it out:

$$\overline A A B + \overline A A \overline C + \overline A\ \overline A \overline B + \overline A \overline B \overline C + A B \overline B + A \overline B \overline C + \overline A \overline B \overline B + \overline B \overline B \overline C + A B C + A \overline C C + \overline A \overline B C + \overline B \overline C C$$

Simplify:

$$\overline A \overline B + \overline B \overline C + ABC$$

$$∑(0,1,4,7) = \overline A \overline B + \overline B \overline C + ABC$$

$$∏(2,3,5,6) = \overline A \overline B + \overline B \overline C + ABC$$

So \$∑(0,1,4,7) = ∏(2,3,5,6)\$.

By a similar process you can show \$∑(0,3,4,6) = ∏(1,2,5,7)\$.

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In Boolean algebra, "product" is normally equivalent to AND, and "sum" is equivalent to OR.

But I'm still not sure how to interpret the question. Obviously, the two lists are complements of each other (when combined, you get 0, 1, 2, 3, 4, 5, 6, 7 without any duplicates, representing all of the possible states of three Boolean variables), but I don't see how that relates to AND and OR.

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The Sigma (capital E looking) signifies a sum, and the pi (i think?) is the product. The SUM is like an add symbol - it provides an OR case. So the ∑(0,3,4,6) means that the output of that circuit is 1 when either 0 OR 3 OR 4 OR 6 are active. So if only 6 was active, the output would be (0+0+0+1) making the output 1, so it works. The Product is like a multiplication, and the same case as an AND gate. So for ∏(1,2,5,7), the circuit would turn on when all of those are activated, i.e. (1 x 1 x 1 x 1 is the only combination that would activate the circuit). You must use this logic to make a circuit that fulfills the question.

All the while, bearing in mind that the 0 to 7 are decimal equivalents of the binary representation of your 3 switches - and this makes sense. 3 input means a total number of 2^3 = 8 combinations - e.g. 000,001,010,011 and so forth.

I guess the question asks to use the logical axioms you know of (should have been taught! there are many important ones) and describe how the circuit that ORs the 0,3,4,6 is the same outputs as one that ANDs 1,2,4,7. Think about shortening it into logical steps that enable you to simplify the equations and make the other one. The question you have been given is basically a proof question - go from one to the other.

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