Sr Examen
Expresión A∨(¬A⇔B)∧C
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Solución
Solución detallada
$$b ⇔ \neg a = \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
$$c \wedge \left(b ⇔ \neg a\right) = c \wedge \left(a \vee b\right) \wedge \left(\neg a \vee \neg b\right)$$
$$a \vee \left(c \wedge \left(b ⇔ \neg a\right)\right) = a \vee \left(b \wedge c\right)$$
$$c \wedge \left(b ⇔ \neg a\right) = c \wedge \left(a \vee b\right) \wedge \left(\neg a \vee \neg b\right)$$
$$a \vee \left(c \wedge \left(b ⇔ \neg a\right)\right) = a \vee \left(b \wedge c\right)$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 0 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 1 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 1 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+