NAND is not associative: (A ↑ B) ↑ C vs A ↑ (B ↑ C) — NAND gate
coredifferent functions, same symbols
Answer
(A ↑ B) ↑ C column 10101011, A ↑ (B ↑ C) column 11110001
Why this example is worth doing
Two bracketings of the same three symbols produce different truth tables, which is not true of AND, OR or XOR. NAND and NOR are the standard counterexamples to the assumption that all binary logic operators associate, and it is an exam favourite. The consequence for design is concrete: a three-input NAND is not two two-input NANDs chained, so a NAND-only conversion has to be done algebraically rather than by textual substitution.
Try your own input in the NAND gate. Truth table, symbol and algebraic form for (A·B)′, the universal gate.
How the answer is reached
Output table
| A | B | C | (A ↑ B) ↑ C | A ↑ (B ↑ C) |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 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 | 1 | 1 |