Q.State any one Distributive Law of Boolean Algebra and verify it using truth table.
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Start your 14-day free trial to unlock the full solution →One Distributive Law of Boolean Algebra: A . (B + C) = A . B + A . C (AND distributes over OR). Building the truth table for all 8 combinations of A, B, C shows the column for A.(B+C) is identical to the column for A.B + A.C, which verifies the law.
Concept — what a Distributive Law says
Boolean Algebra has two distributive laws (each operation distributes over the other):
- AND over OR:
A . (B + C) = A . B + A . C - OR over AND:
A + (B . C) = (A + B) . (A + C)
We state and verify the first one. Here + is OR and . is AND. "Verify using truth table" means: evaluate the left-hand side and the right-hand side independently for every one of the 2^3 = 8 input combinations and show the two result columns match row by row.
Truth-table verification of A . (B + C) = A . B + A . C
| A | B | C | B + C | A . (B + C) (LHS) | A . B | A . C | A.B + A.C (RHS) |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
The LHS column and the RHS column are identical in all 8 rows, so the law holds.
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