Sr Examen
Expresión ¬B~¬A∧B∨C→A∨B
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
(¬b)⇔((c∨(b∧(¬a)))⇒(a∨b))
$$\left(\left(c \vee \left(b \wedge \neg a\right)\right) \Rightarrow \left(a \vee b\right)\right) ⇔ \neg b$$
Solución detallada
$$\left(c \vee \left(b \wedge \neg a\right)\right) \Rightarrow \left(a \vee b\right) = a \vee b \vee \neg c$$
$$\left(\left(c \vee \left(b \wedge \neg a\right)\right) \Rightarrow \left(a \vee b\right)\right) ⇔ \neg b = \neg b \wedge \left(a \vee \neg c\right)$$
$$\left(\left(c \vee \left(b \wedge \neg a\right)\right) \Rightarrow \left(a \vee b\right)\right) ⇔ \neg b = \neg b \wedge \left(a \vee \neg c\right)$$
Tabla de verdad
+---+---+---+--------+ | a | b | c | result | +===+===+===+========+ | 0 | 0 | 0 | 1 | +---+---+---+--------+ | 0 | 0 | 1 | 0 | +---+---+---+--------+ | 0 | 1 | 0 | 0 | +---+---+---+--------+ | 0 | 1 | 1 | 0 | +---+---+---+--------+ | 1 | 0 | 0 | 1 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 0 | +---+---+---+--------+
FND
[src]
$$\left(a \wedge \neg b\right) \vee \left(\neg b \wedge \neg c\right)$$
(a∧(¬b))∨((¬b)∧(¬c))
FNDP
[src]
$$\left(a \wedge \neg b\right) \vee \left(\neg b \wedge \neg c\right)$$
(a∧(¬b))∨((¬b)∧(¬c))