I am having a little trouble knowing what theorem to use to solve the following problem.

Express the given equation without using OR gates. Hint: the expression two NANDed literals looks like this: (X+Y)'


Any help is much appreciated.

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    \$\begingroup\$ Sounds like homework. Maybe you could try making a truth table to see if it looks familiar. \$\endgroup\$ – Elliot Alderson Sep 20 '18 at 1:25
  • \$\begingroup\$ Hi Paul, welcome to EE.SE. Please edit your question and include your attempt to solve the question. \$\endgroup\$ – Hazem Sep 20 '18 at 3:31
  • \$\begingroup\$ Paul, start by first minimizing your equation before worrying about "what theorem" to use. You might already "see" that your equation sums (OR's) three of the possible four ways to combine A and B with AND. That should already give you a clue. Just note that your answer is the NOT of your missing term. The rest should be easy, then. \$\endgroup\$ – jonk Sep 20 '18 at 4:21
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    \$\begingroup\$ I had been trying to use demorgan's theorem, but I was somewhat getting confused between the difference of demorgan's equivalence and demorgan compliments. I did set up a truth table, but was unsure of what to do after this point. If I use the product of sums or sum of products Im still going to have ors. \$\endgroup\$ – Paul Allen Sep 20 '18 at 14:00
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    \$\begingroup\$ @PaulAllen Your own "Hint" in your question suggests a direction for how you might deal with OR situations to make them into AND (or NAND.) So, \$\overline{X+Y}=\overline{X}\cdot\overline{Y}\$ and also \$X+Y=\overline{\overline{X}\cdot\overline{Y}}\$ and also \$X+Y+Z=\overline{\overline{X}\cdot\overline{Y}\cdot\overline{Z}}\$, etc. Again, I think it would be easier if you simplified your equation, first. But you don't have to do so. You can stick with what you have and still use the above to get where you need to go. \$\endgroup\$ – jonk Sep 20 '18 at 17:44

This reads more as a cancellation of terms to me,

when B is "1" both states of A give an output of "1"

when A is "0" both states of B give an output of "1"

So the equation ends up Y'= A*B'


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