Applying De Morgan's to express $pq+r$ in terms of NOR operator

Question

In Boolean Algebras I have $pq+r$ which I think is the same as $(p+r)(q+r)$.

Now, I need to use De Morgan's laws to synthesize this into the NOR form but I am not sure how to apply the laws here.

Answer 1

$\begin{align} pq + r & = (p+r)(q+r)\tag{Distributive Law}\\ \\ & = [(p+r)(q+r)]'' \tag{Double Negation}\\ \\ & = [(p+r)'+ (q+r)']'\tag{DeMorgan's Law}\end{align}$

Can you take it from here? Depending on your notation, the final line of equalities gives us

$$\operatorname{NOR}\Big(\operatorname{NOR}(p, r), \operatorname{NOR}(q, r)\Big)$$