Full subtractor using a 1:4 DEMUX

A full subtractor subtracts $B$ and a borrow-in $B_{in}$ from $A$, giving difference $D$ and borrow-out $B_o$.

Truth table

$$D = \Sigma m(1,2,4,7) = A \oplus B \oplus B_{in}, \qquad B_o = \Sigma m(1,2,3,7)$$

Using a 1:4 DEMUX as a 2-to-4 decoder

Connect the data input of the 1:4 demultiplexer to logic 1 and use A, B as its select lines $S_1S_0$. Each output is then one minterm of A and B:

$$Y_0 = \overline{A}\,\overline{B},\quad Y_1 = \overline{A}B,\quad Y_2 = A\overline{B},\quad Y_3 = AB$$

Group the truth table by A, B and see how D and $B_o$ depend on $B_{in}$:

So:

$$\begin{aligned} D &= B_{in}(Y_0+Y_3) + \overline{B_{in}}(Y_1+Y_2) = B_{in} \oplus (Y_1+Y_2) \\ B_o &= Y_1 + B_{in}(Y_0+Y_3) \end{aligned}$$

(Since exactly one Y is 1, $Y_0+Y_3 = \overline{Y_1+Y_2}$.)

Circuit

        +------------+
 1 ────>| Din    Y0  |── Y0
        |        Y1  |── Y1
        |  1:4   Y2  |── Y2
        | DEMUX  Y3  |── Y3
        +------------+
          S1    S0
          A     B

Y1, Y2 ─────[OR]── Q
Q, Bin ─────[XOR]─────────────── D

Y0, Y3 ─────[OR]── P
P, Bin ─────[AND]── Bin·P ──┐
                            ├─[OR]── Bo
Y1 ─────────────────────────┘

Gates needed: one 1:4 DEMUX, three 2-input OR gates, one AND gate and one XOR gate.

Check: A=0, B=1, $B_{in}$=1 (minterm 3): $Y_1 = 1$, so $Q = 1$, $D = 1 \oplus 1 = 0$ and $B_o = Y_1 = 1$. This matches the truth table.