Realize XNOR Logic Function Using NAND Gates Only

The XNOR gate implements the Boolean function:
$Y = A \odot B = (A\cdot B) + (A' \cdot B')$
Equivalently using XOR and NOT:
$Y = \neg(A \oplus B)$
where $A \oplus B = (A' \cdot B) + (A \cdot B')$.

To “realize using NAND only,” we use the NAND gate as the only primitive. The key identities are:

• Inverter via NAND: $\neg A = \text{NAND}(A,A)$
• NAND is functionally complete: any AND/OR/NOT expression can be rewritten using only NAND with De Morgan’s laws.

We will construct XNOR structurally from NANDs by (1) building XOR using NANDs, then (2) inverting it to get XNOR using one more NAND.

Core NAND-only rewrite rules

Using De Morgan’s laws:

• De Morgan’s law: $(AB)' = A' + B'$

• Thus AND from NAND: $A\cdot B = \big((A\cdot B)'\big)'$ Implementing $((A\cdot B)')$ with NAND and then inverting using a NAND-as-inverter.

• OR from NAND: $A + B = \big((A + B)'\big)'$
  And $(A + B)' = A'\cdot B'$, so: $A + B = \big(A'\cdot B'\big)'$
  Again: generate complements with NAND-inverters, then use NAND(s) to form the complemented AND, then invert.

In practice, to build a gate-level circuit, we minimize inversions and reuse intermediate signals.

A concrete NAND-only XNOR gate-level construction (reusable netlist)

Define:

• $A_ n = \neg A = \text{NAND}(A,A)$
• $B_ n = \neg B = \text{NAND}(B,B)$

XOR structure:

• $T_1 = A_n \cdot B$
  Implement AND from NAND: $T_1 = \text{NAND}(\text{NAND}(A_n,B),\ \text{NAND}(A_n,B))$
• $T_2 = A \cdot B_n$
  $T_2 = \text{NAND}(\text{NAND}(A,B_n),\ \text{NAND}(A,B_n))$
• $X = T_1 + T_2$ (OR from NAND): $X = \text{NAND}(\text{NAND}(T_1,T_1),\ \text{NAND}(T_2,T_2))$
  because $\text{NAND}(T_1,T_1)=\neg T_1$ and NAND of complements gives $\neg(\neg T_1\cdot \neg T_2) = T_1+T_2$.

XNOR:

• $Y = \neg X = \text{NAND}(X,X)$

This is guaranteed correct by the functional equivalences of NAND + De Morgan’s transformations.