I want to convert the logic expression $(a +b) \cdot c \cdot \overline{d}$ into a NOR Normalform. I tried to this by first changing the expression into a DNF and then into the NOR Normalform. Here's my attempt:
$$(a +b) \cdot c \cdot \overline{d} \\ \equiv (a \cdot c \cdot \overline{d}) + (b \cdot c \cdot \overline{d}) \\ \equiv \overline{\overline{(a \cdot c \cdot \overline{d})}} + \overline{\overline{(b \cdot c \cdot \overline{d})}} \\ \equiv \overline{(\overline{a} + \overline{c} + d)} + \overline{(\overline{b} + \overline{c} + d)} \\ \equiv \overline{\overline{\overline{(\overline{a} + \overline{c} + d)} + \overline{(\overline{b} + \overline{c} + d)}}}$$
I am not if this is enough because I still don't think that this is in the NOR form, since the double negation will "cancel out" according to DeMorgan's Law. If I simplify the expression by using DeMorgan's Law, I would get a NAND expression and if I simplify it again I would get negated terms that are not in the NOR form. I'm kind of lost at this point.
Best Answer
$3$-ary DeMorgan's Laws $$\overline{(P\cdot Q\cdot R)}\equiv\overline{P}+\overline{Q}+\overline{R}\tag{1}$$ $$\overline{(P+Q+R)}\equiv\overline{P}\cdot\overline{Q}\cdot\overline{R}\tag{2}$$ Useful links about $n$-ary DeMorgan's Laws:
Apply $3$-ary DeMorgan's Law to the whole term we have \begin{align} (a+b)\cdot c\cdot\overline{d}\equiv&\overline{\overline{(a+b)}+\overline{c}+d}\tag*{DeMorgan's Law $(1)$}\\ \equiv&\overline{\text{Nor}(a,b)+\overline{c}+d}\tag*{Nor definition}\\ \equiv&\text{Nor}(\text{Nor}(a,b),\overline{c},d)\tag*{Nor definition} \end{align} (Using $2$ and $3$ inputs Nor)