F = (A+B)(C+D)E using NOR gates only

$F = (A+B)(C+D)E$ is a product of sums (POS). A POS form maps directly to a NOR-NOR circuit, using double inversion and De Morgan's law.

$$\begin{aligned} F &= (A+B)(C+D)E \\ &= \overline{\overline{(A+B)(C+D)E}} \\ &= \overline{\overline{A+B} + \overline{C+D} + \bar E} \quad (\text{De Morgan}) \end{aligned}$$

So:

• $G_1 = \overline{A + B}$ (2-input NOR)
• $G_2 = \overline{C + D}$ (2-input NOR)
• $G_3 = \overline{E + E} = \bar E$ (NOR used as inverter)
• $F = \overline{G_1 + G_2 + G_3}$ (3-input NOR)

 A --[NOR G1]--(A+B)'--+
 B --|                 |
 C --[NOR G2]--(C+D)'--+--[NOR G4]-- F
 D --|                 |   (3-input)
 E --[NOR G3]-- E' ----+
   (inputs tied)

Total: 4 NOR gates (three 2-input, one 3-input). If only 2-input NORs are allowed, G4 can be replaced by: $G_4' = \overline{G_1 + G_2}$, $G_5 = \overline{G_4' + G_4'}$ (= $G_1 + G_2$), $F = \overline{G_5 + G_3}$, giving 6 gates.

Check: A = 1, B = 0, C = 0, D = 1, E = 1 → $G_1 = 0$, $G_2 = 0$, $G_3 = 0$, F = 1; and $(1)(1)(1) = 1$.