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