Expresión ¬(¬(a|(b⇔c))<-(¬b⊕¬c))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\neg b ⊕ \neg c = \left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right)$$
$$b ⇔ c = \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$a | \left(b ⇔ c\right) = \left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
$$\neg \left(a | \left(b ⇔ c\right)\right) = a \wedge \left(b \vee \neg c\right) \wedge \left(c \vee \neg b\right)$$
$$\left(\neg b ⊕ \neg c\right) \Rightarrow \neg \left(a | \left(b ⇔ c\right)\right) = \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$\left(\neg b ⊕ \neg c\right) \not\Rightarrow \neg \left(a | \left(b ⇔ c\right)\right) = \left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right)$$
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right)$$
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
$$\left(b \vee c\right) \wedge \left(b \vee \neg b\right) \wedge \left(c \vee \neg c\right) \wedge \left(\neg b \vee \neg c\right)$$
(b∨c)∧(b∨(¬b))∧(c∨(¬c))∧((¬b)∨(¬c))
$$\left(b \vee c\right) \wedge \left(\neg b \vee \neg c\right)$$
Ya está reducido a FND
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right)$$
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right)$$