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State and prove De Morgan's theorems
Short answer
First theorem: (A + B)' = A'·B'. Second theorem: (A·B)' = A' + B'. Proof: a truth table shows columns for (A+B)' and A'·B' are identical, and columns for (A·B)' and A'+B' are identical, confirming the complement rules.
State and prove De Morgan's theorems.
Statements
- First theorem: the complement of a sum (OR) equals the product (AND) of the complements. (A + B)' = A' · B'
- Second theorem: the complement of a product equals the sum of the complements. (A · B)' = A' + B'
They extend to any number of variables: (A + B + C)' = A'B'C', (ABC)' = A' + B' + C'.
Proof by truth table
| A | B | A+B | (A+B)' | A' | B' | A'·B' | A·B | (A·B)' | A'+B' |
|---|---|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
The columns (A+B)' and A'·B' are identical, which proves the first theorem. The columns (A·B)' and A'+B' are identical, which proves the second theorem.
Algebraic check of the first theorem
Let X = A + B. Then X + X' = 1 and X·X' = 0. Take Y = A'B'.
- X + Y = A + B + A'B' = (A + A'B') + B = (A + B') + B = A + 1 = 1
- X·Y = (A + B)A'B' = AA'B' + BA'B' = 0 + 0 = 0
So Y is the complement of X, giving (A + B)' = A'B'.
✓ Answer checked by Claude Opus 5.5 (high effort)
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