Applying De Morgans Laws to $a+bc+\overline{a}b\overline{c}d$ in terms of the NOR operator

Question

I need to synthesize $f=a+bc+\overline{a}b\overline{c}d$ into the NOR form.

Can I split this since I know that $a+bc=(a+b)(a+c)=\overline{\overline{a+b}+\overline{a+c}}$?

I'm just not sure how to go about the $\overline{a}b\overline{c}d$ part.

Answer 1

Hint: $\overline{a}b\overline{c}d=(\overline{a}\overline{c})(bd)=\overline{\overline{\overline{a}\overline{c}}+\overline{bd}}=\overline{\overline{\overline{a+c}}+\overline{\overline{\overline{b}+\overline{d}}}}$.

That is, $\overline{a}b\overline{c}d=(\operatorname{not}(a\operatorname{nor}c))\operatorname{nor}(\operatorname{not}((\operatorname{not}b)\operatorname{nor}(\operatorname{not}d)))$. Then implement $\operatorname{not}x$ using $x\operatorname{nor}x$.