Sr Examen
Expresión Bx(A\C)=(BxA)A(Bx(A&C))
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Solución
Ha introducido
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(a∧b∧c∧x)⇔(b∧x∧(a|c))
$$\left(b \wedge x \wedge \left(a | c\right)\right) ⇔ \left(a \wedge b \wedge c \wedge x\right)$$
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Solución detallada
$$a | c = \neg a \vee \neg c$$
$$b \wedge x \wedge \left(a | c\right) = b \wedge x \wedge \left(\neg a \vee \neg c\right)$$
$$\left(b \wedge x \wedge \left(a | c\right)\right) ⇔ \left(a \wedge b \wedge c \wedge x\right) = \neg b \vee \neg x$$
$$b \wedge x \wedge \left(a | c\right) = b \wedge x \wedge \left(\neg a \vee \neg c\right)$$
$$\left(b \wedge x \wedge \left(a | c\right)\right) ⇔ \left(a \wedge b \wedge c \wedge x\right) = \neg b \vee \neg x$$
Tabla de verdad
+---+---+---+---+--------+ | a | b | c | x | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 1 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 1 | +---+---+---+---+--------+ | 0 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 0 | 1 | 1 | 0 | 1 | +---+---+---+---+--------+ | 0 | 1 | 1 | 1 | 0 | +---+---+---+---+--------+ | 1 | 0 | 0 | 0 | 1 | +---+---+---+---+--------+ | 1 | 0 | 0 | 1 | 1 | +---+---+---+---+--------+ | 1 | 0 | 1 | 0 | 1 | +---+---+---+---+--------+ | 1 | 0 | 1 | 1 | 1 | +---+---+---+---+--------+ | 1 | 1 | 0 | 0 | 1 | +---+---+---+---+--------+ | 1 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 0 | 1 | +---+---+---+---+--------+ | 1 | 1 | 1 | 1 | 0 | +---+---+---+---+--------+