Разбираемся:
Давай заполним таблицу истинности для логического выражения ¬(C∧B) ∨ (A∧¬C):
- Строка 1: A=0, B=0, C=0
C∧B = 0, ¬(C∧B) = 1
A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 2: A=0, B=0, C=1
C∧B = 0, ¬(C∧B) = 1
¬C = 0, A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 3: A=0, B=1, C=0
C∧B = 0, ¬(C∧B) = 1
A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 4: A=0, B=1, C=1
C∧B = 1, ¬(C∧B) = 0
¬C = 0, A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 0
- Строка 5: A=1, B=0, C=0
C∧B = 0, ¬(C∧B) = 1
A∧¬C = 1, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 6: A=1, B=0, C=1
C∧B = 0, ¬(C∧B) = 1
¬C = 0, A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 7: A=1, B=1, C=0
C∧B = 0, ¬(C∧B) = 1
A∧¬C = 1, ¬(C∧B) ∨ (A∧¬C) = 1
- Строка 8: A=1, B=1, C=1
C∧B = 1, ¬(C∧B) = 0
¬C = 0, A∧¬C = 0, ¬(C∧B) ∨ (A∧¬C) = 0
| A | B | C | -(C∧B) ∨ (A∧¬C) |
|---|
| 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 |