Sr Examen
Expresión av!c&(a&v!b)=!cva
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Solución
Ha introducido
[src]
(a∨(¬c))⇔(a∨((¬c)∧(a∨(¬b))))
$$\left(a \vee \left(\neg c \wedge \left(a \vee \neg b\right)\right)\right) ⇔ \left(a \vee \neg c\right)$$
Solución detallada
$$a \vee \left(\neg c \wedge \left(a \vee \neg b\right)\right) = a \vee \left(\neg b \wedge \neg c\right)$$
$$\left(a \vee \left(\neg c \wedge \left(a \vee \neg b\right)\right)\right) ⇔ \left(a \vee \neg c\right) = a \vee c \vee \neg b$$
$$\left(a \vee \left(\neg c \wedge \left(a \vee \neg b\right)\right)\right) ⇔ \left(a \vee \neg c\right) = a \vee c \vee \neg b$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 1 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+