Составить таблицу истинности: a)(Av В)& С = (А&О) v (В&О); б)A&B)VCE (AVO&(A&B).

Question

Составить таблицу истинности:
a)(Av В)& С => (А&О) v (В&О);
б)A&B)VCE (AVO&(A&B).

Answer

Конечно! Давайте составим таблицы истинности для предложенных логических выражений.

а) $ (A \vee B) \wedge C \Rightarrow (A \wedge O) \vee (B \wedge O) $

1. Определим переменные $A$, $B$, $C$ и $O$.

2. Рассмотрим все возможные комбинации значений этих переменных (0 - ложь, 1 - истина).

3. Построим таблицу истинности.

A |
B |
C |
O |
A ∨ B |
(A ∨ B) ∧ C |
A ∧ O |
B ∧ O |
(A ∧ O) ∨ (B ∧ O) |
(A ∨ B) ∧ C ⇒ (A ∧ O) ∨ (B ∧ O) |

0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |

0 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |

0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |

0 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |

1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |

1 |
0 |
0 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |

1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
1 |
1 |

1 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |

1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |

б) $ (A \wedge B) \vee C \Rightarrow (A \vee O) \wedge (A \wedge B) $

1. Составим таблицу истинности аналогичным образом.

A |
B |
C |
O |
A ∧ B |
(A ∧ B) ∨ C |
A ∨ O |
A ∧ B |
(A ∨ O) ∧ (A ∧ B) |
(A ∧ B) ∨ C ⇒ (A ∨ O) ∧ (A ∧ B) |

0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |

0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |

0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |

0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |

0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |

0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |

0 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |

1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |

1 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |

1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |

1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |

1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |

1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |

Теперь у нас есть завершенные таблицы истинности для обоих выражений. Если у вас есть еще вопросы или нужна помощь, не стесняйтесь обращаться!

Составить таблицу истинности: a)(Av В)& С => (А&О) v (В&О); б)A&B)VCE (AVO&(A&B).

а) \( (A \vee B) \wedge C \Rightarrow (A \wedge O) \vee (B \wedge O) \)

б) \( (A \wedge B) \vee C \Rightarrow (A \vee O) \wedge (A \wedge B) \)

A	B	C	O	A ∨ B	(A ∨ B) ∧ C	A ∧ O	B ∧ O	(A ∧ O) ∨ (B ∧ O)	(A ∨ B) ∧ C ⇒ (A ∧ O) ∨ (B ∧ O)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	0	0	0	1
0	0	1	0	0	0	0	0	0	1
0	0	1	1	0	0	0	0	0	1
0	1	0	0	1	0	0	0	0	1
0	1	0	1	1	0	0	1	1	1
0	1	1	0	1	1	0	0	0	0
0	1	1	1	1	1	0	1	1	1
1	0	0	0	1	1	1	0	1	1
1	0	0	1	1	1	1	0	1	1
1	0	1	0	1	1	1	0	1	1
1	0	1	1	1	1	1	0	1	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1

A	B	C	O	A ∧ B	(A ∧ B) ∨ C	A ∨ O	A ∧ B	(A ∨ O) ∧ (A ∧ B)	(A ∧ B) ∨ C ⇒ (A ∨ O) ∧ (A ∧ B)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	1	0	0	1
0	0	1	0	0	1	0	0	0	1
0	0	1	1	0	1	1	0	0	1
0	1	0	0	0	0	0	0	0	1
0	1	0	1	0	0	1	0	0	1
0	1	1	0	0	1	0	0	0	1
0	1	1	1	0	1	1	0	0	1
1	0	0	0	0	0	1	0	0	1
1	0	0	1	0	0	1	0	0	1
1	0	1	0	0	1	1	0	0	1
1	0	1	1	0	1	1	0	0	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1

A	B	C	O	A ∨ B	(A ∨ B) ∧ C	A ∧ O	B ∧ O	(A ∧ O) ∨ (B ∧ O)	(A ∨ B) ∧ C ⇒ (A ∧ O) ∨ (B ∧ O)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	0	0	0	1
0	0	1	0	0	0	0	0	0	1
0	0	1	1	0	0	0	0	0	1
0	1	0	0	1	0	0	0	0	1
0	1	0	1	1	0	0	1	1	1
0	1	1	0	1	1	0	0	0	0
0	1	1	1	1	1	0	1	1	1
1	0	0	0	1	1	1	0	1	1
1	0	0	1	1	1	1	0	1	1
1	0	1	0	1	1	1	0	1	1
1	0	1	1	1	1	1	0	1	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1

A	B	C	O	A ∧ B	(A ∧ B) ∨ C	A ∨ O	A ∧ B	(A ∨ O) ∧ (A ∧ B)	(A ∧ B) ∨ C ⇒ (A ∨ O) ∧ (A ∧ B)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	1	0	0	1
0	0	1	0	0	1	0	0	0	1
0	0	1	1	0	1	1	0	0	1
0	1	0	0	0	0	0	0	0	1
0	1	0	1	0	0	1	0	0	1
0	1	1	0	0	1	0	0	0	1
0	1	1	1	0	1	1	0	0	1
1	0	0	0	0	0	1	0	0	1
1	0	0	1	0	0	1	0	0	1
1	0	1	0	0	1	1	0	0	1
1	0	1	1	0	1	1	0	0	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1

A	B	C	O	A ∨ B	(A ∨ B) ∧ C	A ∧ O	B ∧ O	(A ∧ O) ∨ (B ∧ O)	(A ∨ B) ∧ C ⇒ (A ∧ O) ∨ (B ∧ O)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	0	0	0	1
0	0	1	0	0	0	0	0	0	1
0	0	1	1	0	0	0	0	0	1
0	1	0	0	1	0	0	0	0	1
0	1	0	1	1	0	0	1	1	1
0	1	1	0	1	1	0	0	0	0
0	1	1	1	1	1	0	1	1	1
1	0	0	0	1	1	1	0	1	1
1	0	0	1	1	1	1	0	1	1
1	0	1	0	1	1	1	0	1	1
1	0	1	1	1	1	1	0	1	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1

A	B	C	O	A ∧ B	(A ∧ B) ∨ C	A ∨ O	A ∧ B	(A ∨ O) ∧ (A ∧ B)	(A ∧ B) ∨ C ⇒ (A ∨ O) ∧ (A ∧ B)
0	0	0	0	0	0	0	0	0	1
0	0	0	1	0	0	1	0	0	1
0	0	1	0	0	1	0	0	0	1
0	0	1	1	0	1	1	0	0	1
0	1	0	0	0	0	0	0	0	1
0	1	0	1	0	0	1	0	0	1
0	1	1	0	0	1	0	0	0	1
0	1	1	1	0	1	1	0	0	1
1	0	0	0	0	0	1	0	0	1
1	0	0	1	0	0	1	0	0	1
1	0	1	0	0	1	1	0	0	1
1	0	1	1	0	1	1	0	0	1
1	1	0	0	1	1	1	1	1	1
1	1	0	1	1	1	1	1	1	1
1	1	1	0	1	1	1	1	1	1
1	1	1	1	1	1	1	1	1	1