Sr Examen
Expresión ¬B&Av(¬A&B)&(¬BvA)vA&B=B
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
b⇔((a∧b)∨(a∧(¬b))∨(b∧(¬a)∧(a∨(¬b))))
$$b ⇔ \left(\left(a \wedge b\right) \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a \wedge \left(a \vee \neg b\right)\right)\right)$$
Solución detallada
$$b \wedge \neg a \wedge \left(a \vee \neg b\right) = \text{False}$$
$$\left(a \wedge b\right) \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a \wedge \left(a \vee \neg b\right)\right) = a$$
$$b ⇔ \left(\left(a \wedge b\right) \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a \wedge \left(a \vee \neg b\right)\right)\right) = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\left(a \wedge b\right) \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a \wedge \left(a \vee \neg b\right)\right) = a$$
$$b ⇔ \left(\left(a \wedge b\right) \vee \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a \wedge \left(a \vee \neg b\right)\right)\right) = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
Simplificación
[src]
$$\left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
(a∧b)∨((¬a)∧(¬b))
Tabla de verdad
+---+---+--------+ | a | b | result | +===+===+========+ | 0 | 0 | 1 | +---+---+--------+ | 0 | 1 | 0 | +---+---+--------+ | 1 | 0 | 0 | +---+---+--------+ | 1 | 1 | 1 | +---+---+--------+
FND
[src]
Ya está reducido a FND
$$\left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
(a∧b)∨((¬a)∧(¬b))
FNC
[src]
$$\left(a \vee \neg a\right) \wedge \left(a \vee \neg b\right) \wedge \left(b \vee \neg a\right) \wedge \left(b \vee \neg b\right)$$
(a∨(¬a))∧(a∨(¬b))∧(b∨(¬a))∧(b∨(¬b))