2-bit equality comparator

Two 2-bit numbers $A = A_1A_0$ and $B = B_1B_0$ are equal only when each pair of corresponding bits is equal. Bit equality is given by the XNOR gate.

Truth table (16 rows; x = 1 only when A = B)

So

$$x = A_1'A_0'B_1'B_0' + A_1'A_0B_1'B_0 + A_1A_0'B_1B_0' + A_1A_0B_1B_0$$

Simplification

Group the terms by the bit-1 pair and the bit-0 pair:

$$\begin{aligned} x &= (A_1'B_1' + A_1B_1)(A_0'B_0' + A_0B_0) \\ &= (A_1 \odot B_1)(A_0 \odot B_0) \\ &= \overline{(A_1 \oplus B_1)}\cdot\overline{(A_0 \oplus B_0)} \end{aligned}$$

The K-map (rows $A_1A_0$, columns $B_1B_0$) has 1s only on the diagonal cells 0, 5, 10, 15, which cannot be grouped, confirming that the SOP form has four terms; the XNOR form is the simplest circuit.

Logic circuit

 A1 --+
      +--[XNOR]-- E1 --+
 B1 --+                |
                       +--[AND]--- x
 A0 --+                |
      +--[XNOR]-- E0 --+
 B0 --+

Alternative: $x = \overline{(A_1 \oplus B_1) + (A_0 \oplus B_0)}$, i.e. two XOR gates followed by one NOR gate.

Check: $A = 10$, $B = 10$: $E_1 = 1$, $E_0 = 1$, so $x = 1$. $A = 10$, $B = 11$: $E_0 = 0$, so $x = 0$.