Sr Examen
Expresión a&(¬avb)⇒(a&c⇔¬b)
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Solución
Ha introducido
[src]
(a∧(b∨(¬a)))⇒((¬b)⇔(a∧c))
$$\left(a \wedge \left(b \vee \neg a\right)\right) \Rightarrow \left(\left(a \wedge c\right) ⇔ \neg b\right)$$
Solución detallada
$$a \wedge \left(b \vee \neg a\right) = a \wedge b$$
$$\left(a \wedge c\right) ⇔ \neg b = \left(b \wedge \neg a\right) \vee \left(b \wedge \neg c\right) \vee \left(a \wedge c \wedge \neg b\right)$$
$$\left(a \wedge \left(b \vee \neg a\right)\right) \Rightarrow \left(\left(a \wedge c\right) ⇔ \neg b\right) = \neg a \vee \neg b \vee \neg c$$
$$\left(a \wedge c\right) ⇔ \neg b = \left(b \wedge \neg a\right) \vee \left(b \wedge \neg c\right) \vee \left(a \wedge c \wedge \neg b\right)$$
$$\left(a \wedge \left(b \vee \neg a\right)\right) \Rightarrow \left(\left(a \wedge c\right) ⇔ \neg b\right) = \neg a \vee \neg b \vee \neg c$$
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 | 0 | +---+---+---+--------+