K-map Σm(7,9,12,13,14,15)+d, NOR-only

Given $F(A,B,C,D) = \Sigma m(7,9,12,13,14,15) + d(0,2,3,5)$.

K-map simplification

        CD
AB      00  01  11  10
      +---+---+---+---+
 00   | X | 0 | X | X |
      +---+---+---+---+
 01   | 0 | X | 1 | 0 |
      +---+---+---+---+
 11   | 1 | 1 | 1 | 1 |
      +---+---+---+---+
 10   | 0 | 1 | 0 | 0 |
      +---+---+---+---+

Groups (don't-care 5 is used in the BD quad; 0, 2, 3 are not needed):

• Quad m(12,13,14,15) → $AB$
• Pair m(9,13) → $A\overline{C}D$
• Quad m(5,7,13,15) → $BD$

$$F = AB + BD + A\overline{C}D$$

Check: 12, 13, 14, 15 by AB; 7 (and 5, 13, 15) by BD; 9 by $A\overline{C}D$.

Circuit using NOR gates only

For NOR-only realization of an SOP expression, write each product term as a NOR of complemented literals (De Morgan: $\overline{X}\,\overline{Y} = \overline{X+Y}$), feed these to a NOR gate (which gives $\overline{F}$), and invert with one more NOR whose inputs are tied.

$$\begin{aligned} AB &= \overline{\overline{A}+\overline{B}} \\ BD &= \overline{\overline{B}+\overline{D}} \\ A\overline{C}D &= \overline{\overline{A}+C+\overline{D}} \\ F &= \overline{\overline{P_1+P_2+P_3}} \end{aligned}$$

A ──[NOR]── A'   (inputs tied)
B ──[NOR]── B'   (inputs tied)
D ──[NOR]── D'   (inputs tied)

A', B'    ──[NOR]── P1  ─┐
B', D'    ──[NOR]── P2  ─┤──[NOR]── F'
A', C, D' ──[NOR]── P3  ─┘

F' ──[NOR]── F   (inputs tied)

Gates: 3 NOR inverters ($\overline{A}, \overline{B}, \overline{D}$), 3 first-level NORs, a 3-input NOR giving $\overline{F}$ and a final NOR inverter.