Sr Examen
Expresión ¬(x3∧¬x2)⇒(x3∨¬x2∨x1)⇔x3
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
x3⇔((¬(x3∧(¬x2)))⇒(x1∨x3∨(¬x2)))
$$x_{3} ⇔ \left(\neg \left(x_{3} \wedge \neg x_{2}\right) \Rightarrow \left(x_{1} \vee x_{3} \vee \neg x_{2}\right)\right)$$
Solución detallada
$$\neg \left(x_{3} \wedge \neg x_{2}\right) = x_{2} \vee \neg x_{3}$$
$$\neg \left(x_{3} \wedge \neg x_{2}\right) \Rightarrow \left(x_{1} \vee x_{3} \vee \neg x_{2}\right) = x_{1} \vee x_{3} \vee \neg x_{2}$$
$$x_{3} ⇔ \left(\neg \left(x_{3} \wedge \neg x_{2}\right) \Rightarrow \left(x_{1} \vee x_{3} \vee \neg x_{2}\right)\right) = x_{3} \vee \left(x_{2} \wedge \neg x_{1}\right)$$
$$\neg \left(x_{3} \wedge \neg x_{2}\right) \Rightarrow \left(x_{1} \vee x_{3} \vee \neg x_{2}\right) = x_{1} \vee x_{3} \vee \neg x_{2}$$
$$x_{3} ⇔ \left(\neg \left(x_{3} \wedge \neg x_{2}\right) \Rightarrow \left(x_{1} \vee x_{3} \vee \neg x_{2}\right)\right) = x_{3} \vee \left(x_{2} \wedge \neg x_{1}\right)$$
Tabla de verdad
+----+----+----+--------+ | x1 | x2 | x3 | result | +====+====+====+========+ | 0 | 0 | 0 | 0 | +----+----+----+--------+ | 0 | 0 | 1 | 1 | +----+----+----+--------+ | 0 | 1 | 0 | 1 | +----+----+----+--------+ | 0 | 1 | 1 | 1 | +----+----+----+--------+ | 1 | 0 | 0 | 0 | +----+----+----+--------+ | 1 | 0 | 1 | 1 | +----+----+----+--------+ | 1 | 1 | 0 | 0 | +----+----+----+--------+ | 1 | 1 | 1 | 1 | +----+----+----+--------+
FNCD
[src]
$$\left(x_{2} \vee x_{3}\right) \wedge \left(x_{3} \vee \neg x_{1}\right)$$
(x2∨x3)∧(x3∨(¬x1))
FNC
[src]
$$\left(x_{2} \vee x_{3}\right) \wedge \left(x_{3} \vee \neg x_{1}\right)$$
(x2∨x3)∧(x3∨(¬x1))