NOR is not associative: (A ↓ B) ↓ C vs A ↓ (B ↓ C) — NOR gate
coredifferent functions, same symbols
Answer
(A ↓ B) ↓ C column 00101010, A ↓ (B ↓ C) column 01110000
Why this example is worth doing
Like NAND, NOR fails to associate, and the two output columns differ. What this page adds beyond the NAND statement of the same fact is the reason: associativity survives inversion only when the inversion can be pushed outside the whole expression, and a bubble on every intermediate output cannot be. That framing generalises — it tells you in advance which derived operators will and will not associate, rather than requiring a table each time.
Try your own input in the NOR gate. Truth table, symbol and algebraic form for (A + B)′, the other universal gate.
How the answer is reached
Output table
| A | B | C | (A ↓ B) ↓ C | A ↓ (B ↓ C) |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 |