
Составьте таблицу истинности для логического выражения:
1) $\overline{A} ⇒ B$;
2) (A ∨ B) ∧ C;
3) (A ∧ B) ⇒ C;
4) (A ⇒ B) ∧ (B ∨ C);
5) $(A ∧ \overline{C}) ⇒ B$.

$\overline{A} ⇒ B$
| A | $\overline{A}$ | B | $\overline{A} ⇒ B$ |
|---|---|---|---|
| 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 |
(A ∨ B) ∧ C
| A | B | C | A ∨ B | (A ∨ B) ∧ C |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 0 | 0 | 0 |
(A ∧ B) ⇒ C
| A | B | C | A ∧ B | (A ∧ B) ⇒ C |
|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 |
(A ⇒ B) ∧ (B ∨ C)
| A | B | C | A ⇒ B | B ∨ C | (A ⇒ B) ∧ (B ∨ C) |
|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 |
$(A ∧ \overline{C}) ⇒ B$
| A | B | C | $\overline{C}$ | $A ∧ \overline{C}$ | $(A ∧ \overline{C}) ⇒ B$ |
|---|---|---|---|---|---|
| 1 | 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 | 1 |
Пожалуйста, оцените решение