Краткое пояснение: Заполняем таблицу истинности, вычисляя значения логических выражений для каждой комбинации входных данных L, B, C.
Пошаговое решение:
Смотри, как это работает:
- Строка 1: L=0, B=0, C=0
- L ∧ C = 0 ∧ 0 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 0 ∧ 1 = 0
- G = L ∨ B ∧ ¬(L ∧ C) = 0 ∨ 0 = 0
- Строка 2: L=0, B=0, C=1
- L ∧ C = 0 ∧ 1 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 0 ∧ 1 = 0
- G = L ∨ B ∧ ¬(L ∧ C) = 0 ∨ 0 = 0
- Строка 3: L=0, B=1, C=0
- L ∧ C = 0 ∧ 0 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 1 ∧ 1 = 1
- G = L ∨ B ∧ ¬(L ∧ C) = 0 ∨ 1 = 1
- Строка 4: L=0, B=1, C=1
- L ∧ C = 0 ∧ 1 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 1 ∧ 1 = 1
- G = L ∨ B ∧ ¬(L ∧ C) = 0 ∨ 1 = 1
- Строка 5: L=1, B=0, C=0
- L ∧ C = 1 ∧ 0 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 0 ∧ 1 = 0
- G = L ∨ B ∧ ¬(L ∧ C) = 1 ∨ 0 = 1
- Строка 6: L=1, B=0, C=1
- L ∧ C = 1 ∧ 1 = 1
- ¬(L ∧ C) = ¬1 = 0
- B ∧ ¬(L ∧ C) = 0 ∧ 0 = 0
- G = L ∨ B ∧ ¬(L ∧ C) = 1 ∨ 0 = 1
- Строка 7: L=1, B=1, C=0
- L ∧ C = 1 ∧ 0 = 0
- ¬(L ∧ C) = ¬0 = 1
- B ∧ ¬(L ∧ C) = 1 ∧ 1 = 1
- G = L ∨ B ∧ ¬(L ∧ C) = 1 ∨ 1 = 1
- Строка 8: L=1, B=1, C=1
- L ∧ C = 1 ∧ 1 = 1
- ¬(L ∧ C) = ¬1 = 0
- B ∧ ¬(L ∧ C) = 1 ∧ 0 = 0
- G = L ∨ B ∧ ¬(L ∧ C) = 1 ∨ 0 = 1
Ответ:
| L |
B |
C |
L ∧ C |
¬(L ∧ C) |
B ∧ ¬(L ∧ C) |
G |
| 0 |
0 |
0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
0 |
1 |
0 |
0 |
| 0 |
1 |
0 |
0 |
1 |
1 |
1 |
| 0 |
1 |
1 |
0 |
1 |
1 |
1 |
| 1 |
0 |
0 |
0 |
1 |
0 |
1 |
| 1 |
0 |
1 |
1 |
0 |
0 |
1 |
| 1 |
1 |
0 |
0 |
1 |
1 |
1 |
| 1 |
1 |
1 |
1 |
0 |
0 |
1 |