# How do I make this logic circuit (A+B)' (B+D)' using NAND gates only?

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

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

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