What I have done so far:

\begin{align*} (A + B)'(B + D)' = A'B'B'D' = (A + B + D)' \end{align*}

I always get stuck at this point, I don't know what to do next that would allow me to use the NAND gates.

I would appreciate any hints or answers

  • $\begingroup$ If the prime means not then isn't $(A+B)'$ already a nand gate? $\endgroup$
    – John Douma
    Jul 2 at 16:15
  • $\begingroup$ You could use that A' = (AA)' and then A+B = (A' B')' $\endgroup$
    – S.Klumpers
    Jul 2 at 16:28
  • $\begingroup$ A'B'D'=nand(nand(A'B',D'),nand(A'B',D')) where A'B'=nand(nand(A',B'),nand(A',B')) $\endgroup$
    – Manx
    Jul 2 at 18:43
  • 1
    $\begingroup$ @JohnDouma $(A+B)'$ is a NOR gate since $+$ means OR. $\endgroup$
    – bobeyt6
    Jul 2 at 19:28

1 Answer 1


If we start from your last relation: \begin{align} (A+B+D)'&=A'B'D'\\&=\left(\left(A'B'D'\right)'\right)' \end{align} Here, we have only NAND gates. $A'$ can be implemented as a NAND gate (with $A$ and $A$ as input signals). To build the 3-way NAND from 2-way NAND gates, you can do something like: https://electronics.stackexchange.com/questions/211756/how-to-build-a-3-input-nand-gate-from-2-input-nand-gates-or-a-3-input-nor-gate-f

So you need 7 gates here (assuming 2-way gates), maybe there are other solutions with fewer gates.


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