Sr Examen
Expresión ¬(¬A&BvA&(Bv¬C))⇔¬B&(¬AvC)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((¬b)∧(c∨(¬a)))⇔(¬((b∧(¬a))∨(a∧(b∨(¬c)))))
$$\left(\neg b \wedge \left(c \vee \neg a\right)\right) ⇔ \neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right)$$
Solución detallada
$$\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right) = b \vee \left(a \wedge \neg c\right)$$
$$\neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right) = \neg b \wedge \left(c \vee \neg a\right)$$
$$\left(\neg b \wedge \left(c \vee \neg a\right)\right) ⇔ \neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right) = 1$$
$$\neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right) = \neg b \wedge \left(c \vee \neg a\right)$$
$$\left(\neg b \wedge \left(c \vee \neg a\right)\right) ⇔ \neg \left(\left(a \wedge \left(b \vee \neg c\right)\right) \vee \left(b \wedge \neg a\right)\right) = 1$$
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 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+