3-input XNOR, A = 0, B = 0, C = 1 — XNOR gate
coreparity, not all-equal
Answer
output 1
Why this example is worth doing
The trap on this page, and a sharper one than the XOR version. Two inputs low and one high are plainly not all equal, yet the chained XNOR outputs 1 — because a chain of XNORs computes parity, with the accumulated inversions deciding which parity, and not agreement. An all-equal detector over three inputs is a different circuit entirely, which the page draws beside this one so the two are never confused again.
Try your own input in the XNOR gate. Truth table, symbol and algebraic form for A ⊙ B, the equality detector.
How the answer is reached
The probed row
A = 0, B = 0, C = 11— Row 1 of 8.
Truth table
| # | A | B | C | A ⊙ B ⊙ C |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 1 |
| 2 | 0 | 1 | 0 | 1 |
| 3 | 0 | 1 | 1 | 0 |
| 4 | 1 | 0 | 0 | 1 |
| 5 | 1 | 0 | 1 | 0 |
| 6 | 1 | 1 | 0 | 0 |
| 7 | 1 | 1 | 1 | 1 |