Sr Examen
Expresión BC+A(!C)=(!A)BC+AB(!C)
El profesor se sorprenderá mucho al ver tu solución correcta😉
⌨
Solución
Ha introducido
[src]
((b∧c)∨(a∧(¬c)))⇔((a∧b∧(¬c))∨(b∧c∧(¬a)))
$$\left(\left(a \wedge \neg c\right) \vee \left(b \wedge c\right)\right) ⇔ \left(\left(a \wedge b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)\right)$$
Solución detallada
$$\left(a \wedge b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right) = b \wedge \left(a \vee c\right) \wedge \left(\neg a \vee \neg c\right)$$
$$\left(\left(a \wedge \neg c\right) \vee \left(b \wedge c\right)\right) ⇔ \left(\left(a \wedge b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)\right) = \left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
$$\left(\left(a \wedge \neg c\right) \vee \left(b \wedge c\right)\right) ⇔ \left(\left(a \wedge b \wedge \neg c\right) \vee \left(b \wedge c \wedge \neg a\right)\right) = \left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
Simplificación
[src]
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
(¬a)∨(b∧(¬c))∨(c∧(¬b))
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 1 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FNDP
[src]
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
(¬a)∨(b∧(¬c))∨(c∧(¬b))
FND
[src]
Ya está reducido a FND
$$\left(b \wedge \neg c\right) \vee \left(c \wedge \neg b\right) \vee \neg a$$
(¬a)∨(b∧(¬c))∨(c∧(¬b))
FNCD
[src]
$$\left(b \vee c \vee \neg a\right) \wedge \left(\neg a \vee \neg b \vee \neg c\right)$$
(b∨c∨(¬a))∧((¬a)∨(¬b)∨(¬c))
FNC
[src]
$$\left(b \vee c \vee \neg a\right) \wedge \left(b \vee \neg a \vee \neg b\right) \wedge \left(c \vee \neg a \vee \neg c\right) \wedge \left(\neg a \vee \neg b \vee \neg c\right)$$
(b∨c∨(¬a))∧(b∨(¬a)∨(¬b))∧(c∨(¬a)∨(¬c))∧((¬a)∨(¬b)∨(¬c))