Seven-segment "e" segment

For a BCD-to-seven-segment decoder the input is a BCD digit $ABCD$ ($A$ = MSB). Codes 1010–1111 never occur, so they are don't cares. Segment e is the lower-left segment (common-cathode display: 1 = ON).

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

Truth table for e

Segment e is ON only for digits 0, 2, 6 and 8.

$$e = \Sigma m(0,2,6,8) + d(10,11,12,13,14,15)$$

K-map

        CD=00 01  11  10
AB=00     1   0   0   1
AB=01     0   0   0   1
AB=11     X   X   X   X
AB=10     1   0   X   X

• Quad 1: the four corners $m_0, m_2, m_8, m_{10}$ give $B'D'$.
• Quad 2: column $CD = 10$: $m_2, m_6, m_{14}, m_{10}$ give $CD'$.

$$e = B'D' + CD' = D'(B' + C)$$

Simplest circuit

The factored form needs only one OR, one AND and two inverters:

B --[NOT]-- B' --+
                 +--[OR]--+
C ---------------+        +--[AND]--> e
D --[NOT]-- D' -----------+

Check: digit 2 (0010): $D' = 1$, $B' + C = 1$, so $e = 1$; digit 4 (0100): $B' = 0$, $C = 0$, so $e = 0$. Both correct.