Minimal SOP Σm(3,4,6,8,10,15)+d, NOR-only

Given $F(A,B,C,D) = \Sigma m(3,4,6,8,10,15) + d(0,2,7,14)$.

K-map and minimal SOP

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

Every group below is a quad, made possible by the don't-cares 0, 2, 7, 14:

• Quad m(2,3,6,7) → $\overline{A}C$
• Quad m(0,2,4,6) → $\overline{A}\,\overline{D}$
• Quad m(0,2,8,10) → $\overline{B}\,\overline{D}$
• Quad m(6,7,14,15) → $BC$

$$F = \overline{A}C + \overline{A}\,\overline{D} + \overline{B}\,\overline{D} + BC$$

Check: 3 (and 2, 7) by $\overline{A}C$; 4, 6 by $\overline{A}\,\overline{D}$; 8, 10 by $\overline{B}\,\overline{D}$; 15 (with 7, 14) by $BC$. This is the only minimal SOP (4 terms, 8 literals).

Implementation 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.

$$\begin{aligned} \overline{A}C &= \overline{A+\overline{C}} & \overline{A}\,\overline{D} &= \overline{A+D} \\ \overline{B}\,\overline{D} &= \overline{B+D} & BC &= \overline{\overline{B}+\overline{C}} \end{aligned}$$ $$F = \overline{\overline{P_1+P_2+P_3+P_4}}$$

B ──[NOR]── B'   (inputs tied)
C ──[NOR]── C'   (inputs tied)

A, C'  ──[NOR]── P1  ─┐
A, D   ──[NOR]── P2  ─┤
                      ├──[NOR]── F'
B, D   ──[NOR]── P3  ─┤
B', C' ──[NOR]── P4  ─┘

F' ──[NOR]── F   (inputs tied)

Gates: 2 NOR inverters ($\overline{B}, \overline{C}$), 4 two-input NORs, one 4-input NOR and one NOR inverter: 8 NOR gates.