Expresión ABC+AB'C+AB'C'+A'BC+A'B'C'
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Solución
Solución detallada
$$\neg \left(a \vee \left(b \wedge c\right)\right) = \neg a \wedge \left(\neg b \vee \neg c\right)$$
$$\neg a \wedge \neg b \wedge \neg c \wedge \neg \left(a \vee \left(b \wedge c\right)\right) = \neg a \wedge \neg b \wedge \neg c$$
$$\neg \left(c \vee \left(a \wedge b\right)\right) = \neg c \wedge \left(\neg a \vee \neg b\right)$$
$$\left(a \wedge b\right) \vee \left(a \wedge b \wedge c\right) = a \wedge b$$
$$\neg \left(\left(a \wedge b\right) \vee \left(a \wedge b \wedge c\right)\right) = \neg a \vee \neg b$$
$$\neg c \wedge \neg \left(c \vee \left(a \wedge b\right)\right) \wedge \neg \left(\left(a \wedge b\right) \vee \left(a \wedge b \wedge c\right)\right) = \neg c \wedge \left(\neg a \vee \neg b\right)$$
$$\left(\neg c \wedge \neg \left(c \vee \left(a \wedge b\right)\right) \wedge \neg \left(\left(a \wedge b\right) \vee \left(a \wedge b \wedge c\right)\right)\right) \vee \left(\neg a \wedge \neg b \wedge \neg c \wedge \neg \left(a \vee \left(b \wedge c\right)\right)\right) = \neg c \wedge \left(\neg a \vee \neg b\right)$$
$$\neg c \wedge \left(\neg a \vee \neg b\right)$$
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 0 |
+---+---+---+--------+
Ya está reducido a FNC
$$\neg c \wedge \left(\neg a \vee \neg b\right)$$
$$\left(\neg a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right)$$
$$\neg c \wedge \left(\neg a \vee \neg b\right)$$
$$\left(\neg a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right)$$