Выражение ¬(x&¬t&y)⇒y&x&¬(x&z)

В калькуляторе вы сможете упростить выражения, содержащие следующие операции: NOT, XOR, AND, OR, NAND, NOR, NOT, XNOR

Вы ввели

(¬(x∧y∧(¬t)))⇒(x∧y∧(¬(x∧z)))

$$\neg \left(x \wedge y \wedge \neg t\right) \Rightarrow \left(x \wedge y \wedge \neg \left(x \wedge z\right)\right)$$

Подробное решение

$$\neg \left(x \wedge y \wedge \neg t\right) = t \vee \neg x \vee \neg y$$
$$\neg \left(x \wedge z\right) = \neg x \vee \neg z$$
$$x \wedge y \wedge \neg \left(x \wedge z\right) = x \wedge y \wedge \neg z$$
$$\neg \left(x \wedge y \wedge \neg t\right) \Rightarrow \left(x \wedge y \wedge \neg \left(x \wedge z\right)\right) = x \wedge y \wedge \left(\neg t \vee \neg z\right)$$

Упрощение

$$x \wedge y \wedge \left(\neg t \vee \neg z\right)$$

Таблица истинности

+---+---+---+---+--------+
| t | x | y | z | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+

СДНФ

$$\left(x \wedge y \wedge \neg t\right) \vee \left(x \wedge y \wedge \neg z\right)$$

КНФ

Уже приведено к КНФ

$$x \wedge y \wedge \left(\neg t \vee \neg z\right)$$

СКНФ

$$x \wedge y \wedge \left(\neg t \vee \neg z\right)$$

ДНФ

$$\left(x \wedge y \wedge \neg t\right) \vee \left(x \wedge y \wedge \neg z\right)$$