Sr Examen
Expresión (¬a+¬b⇔¬c)⇒(a*b⇔c)
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Solución
Ha introducido
[src]
((¬c)⇔((¬a)∨(¬b)))⇒(c⇔(a∧b))
$$\left(\neg c ⇔ \left(\neg a \vee \neg b\right)\right) \Rightarrow \left(c ⇔ \left(a \wedge b\right)\right)$$
Solución detallada
$$\neg c ⇔ \left(\neg a \vee \neg b\right) = \left(\neg a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(a \wedge b \wedge c\right)$$
$$c ⇔ \left(a \wedge b\right) = \left(\neg a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(a \wedge b \wedge c\right)$$
$$\left(\neg c ⇔ \left(\neg a \vee \neg b\right)\right) \Rightarrow \left(c ⇔ \left(a \wedge b\right)\right) = 1$$
$$c ⇔ \left(a \wedge b\right) = \left(\neg a \wedge \neg c\right) \vee \left(\neg b \wedge \neg c\right) \vee \left(a \wedge b \wedge c\right)$$
$$\left(\neg c ⇔ \left(\neg a \vee \neg b\right)\right) \Rightarrow \left(c ⇔ \left(a \wedge b\right)\right) = 1$$
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 | +---+---+---+--------+