Full subtractor from half subtractors

A full subtractor subtracts $B$ and a borrow-in $B_{in}$ from $A$, giving difference $D$ and borrow-out $B_o$. A half subtractor subtracts $Y$ from $X$: $D = X \oplus Y$, $B = X'Y$.

Truth table and expressions

$$D = A \oplus B \oplus B_{in}$$ $$\begin{aligned} B_o &= A'B'B_{in} + A'BB_{in}' + A'BB_{in} + ABB_{in} \\ &= A'B(B_{in}' + B_{in}) + B_{in}(A'B' + AB) \\ &= A'B + (A \oplus B)'\,B_{in} \end{aligned}$$

Using two half subtractors and one OR gate

• HS1 with inputs $A$, $B$: $D_1 = A \oplus B$, $B_1 = A'B$.
• HS2 with inputs $D_1$, $B_{in}$: $D = D_1 \oplus B_{in} = A \oplus B \oplus B_{in}$, $B_2 = D_1'B_{in} = (A \oplus B)'B_{in}$.
• OR gate: $B_o = B_1 + B_2 = A'B + (A \oplus B)'B_{in}$, which matches the expression above.

        +-----+  D1   +-----+
A ----->| HS1 |------>| HS2 |----------> D
B ----->|     |  Bin->|     |
        +-----+       +-----+
           | B1          | B2
           +---->[OR]<---+
                   |
                   +-----------------> Bo

Check: $A=0, B=1, B_{in}=1$: HS1 gives $D_1 = 1$, $B_1 = 1$; HS2 gives $D = 0$, $B_2 = 0$; $B_o = 1$. So $0 - 1 - 1 = -2$, i.e. $D = 0$ with borrow 1. Correct.