Seven-segment "b" and "d" segments

A BCD-to-seven-segment decoder takes BCD input $ABCD$ ($A$ = MSB) and drives segments a–g. Inputs 10–15 never occur, so they are don't-cares (X).

     a
    ---
 f | g | b
    ---
 e |   | c
    ---
     d

Segment "b"

Segment b is ON for digits 0, 1, 2, 3, 4, 7, 8, 9 (OFF for 5, 6).

$$b = \Sigma m(0,1,2,3,4,7,8,9) + d(10\text{–}15)$$

Groups: $B'$ (rows 00 and 10), $C'D'$ (column 00), $CD$ (column 11).

$$b = B' + C'D' + CD = B' + (C \odot D)$$

 B ---[NOT]---------------+
 C --+                    +--[OR]--- b
 D --+--[XNOR]------------+

Segment "d"

Segment d is ON for digits 0, 2, 3, 5, 6, 8, 9 (OFF for 1, 4, 7).

$$d = \Sigma m(0,2,3,5,6,8,9) + d(10\text{–}15)$$

Groups: $A$ (rows 11, 10); $B'D'$ (corners 0, 2, 8, 10); $B'C$ (2, 3, 10, 11); $CD'$ (2, 6, 10, 14); $BC'D$ (5 with 13).

$$d = A + B'D' + B'C + CD' + BC'D$$

 B',D' ---[AND]---+
 B',C  ---[AND]---+
 C,D'  ---[AND]---+--[OR 5]--- d
 B,C',D --[AND]---+
 A ---------------+

Check: digit 5 ($0101$): $b = 0 + 0 + 0 = 0$ (b off) and $d = BC'D = 1$ (d on). Digit 7 ($0111$): $b = CD = 1$ and every term of $d$ is 0. Both expressions were verified for digits 0–9.