Пошаговое решение:
Внимательно проанализируем логическое выражение \( A \lor B \land
eg(A \lor
eg C) \) для каждой строки таблицы истинности.
- Строка 1: A=0, B=0, C=0\(\rightarrow
eg C = 1\) \(\rightarrow A \lor B = 0\) \(\rightarrow A \lor
eg C = 1 \rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 0 \land 0 = 0 \rightarrow A \lor 0 = 0 \) - Строка 2: A=0, B=0, C=1 \(\rightarrow
eg C = 0\) \(\rightarrow A \lor B = 0 \rightarrow A \lor
eg C = 0\rightarrow
eg(A \lor
eg C) = 1\) \(\rightarrow B \land
eg(A \lor
eg C) = 0 \land 1 = 0\rightarrow A \lor 0 = 0 \) - Строка 3: A=0, B=1, C=0\(\rightarrow
eg C = 1\) \(\rightarrow A \lor B = 1\) \(\rightarrow A \lor
eg C = 1 \rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 1 \land 0 = 0\rightarrow A \lor 0 = 0 \) - Строка 4: A=0, B=1, C=1 \(\rightarrow
eg C = 0\) \(\rightarrow A \lor B = 1\) \(\rightarrow A \lor
eg C = 0 \rightarrow
eg(A \lor
eg C) = 1\) \(\rightarrow B \land
eg(A \lor
eg C) = 1 \land 1 = 1 \rightarrow A \lor 1 = 1\) - Строка 5: A=1, B=0, C=0\(\rightarrow
eg C = 1\) \(\rightarrow A \lor B = 1 \rightarrow A \lor
eg C = 1\rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 0 \land 0 = 0\rightarrow A \lor 0 = 1 \) - Строка 6: A=1, B=0, C=1\(\rightarrow
eg C = 0\) \(\rightarrow A \lor B = 1 \rightarrow A \lor
eg C = 1\rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 0 \land 0 = 0\rightarrow A \lor 0 = 1 \) - Строка 7: A=1, B=1, C=0\(\rightarrow
eg C = 1\) \(\rightarrow A \lor B = 1 \rightarrow A \lor
eg C = 1\rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 1 \land 0 = 0\rightarrow A \lor 0 = 1 \) - Строка 8: A=1, B=1, C=1\(\rightarrow
eg C = 0\) \(\rightarrow A \lor B = 1\) \(\rightarrow A \lor
eg C = 1\rightarrow
eg(A \lor
eg C) = 0\) \(\rightarrow B \land
eg(A \lor
eg C) = 1 \land 0 = 0\rightarrow A \lor 0 = 1 \)
Ответ: 0, 0, 0, 1, 1, 1, 1, 1