Sr Examen
Expresión DC’B’A+DC’BA
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
a∧b∧(¬b)∧(¬(c∧d))∧(¬(a∨(c∧d)))
$$a \wedge b \wedge \neg b \wedge \neg \left(c \wedge d\right) \wedge \neg \left(a \vee \left(c \wedge d\right)\right)$$
Solución detallada
$$\neg \left(c \wedge d\right) = \neg c \vee \neg d$$
$$\neg \left(a \vee \left(c \wedge d\right)\right) = \neg a \wedge \left(\neg c \vee \neg d\right)$$
$$a \wedge b \wedge \neg b \wedge \neg \left(c \wedge d\right) \wedge \neg \left(a \vee \left(c \wedge d\right)\right) = \text{False}$$
$$\neg \left(a \vee \left(c \wedge d\right)\right) = \neg a \wedge \left(\neg c \vee \neg d\right)$$
$$a \wedge b \wedge \neg b \wedge \neg \left(c \wedge d\right) \wedge \neg \left(a \vee \left(c \wedge d\right)\right) = \text{False}$$
Tabla de verdad
+---+---+---+---+--------+ | a | b | c | d | result | +===+===+===+===+========+ | 0 | 0 | 0 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 0 | 1 | 0 | +---+---+---+---+--------+ | 0 | 0 | 1 | 0 | 0 | +---+---+---+---+--------+ | 0 | 0 | 1 | 1 | 0 | +---+---+---+---+--------+ | 0 | 1 | 0 | 0 | 0 | +---+---+---+---+--------+ | 0 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 0 | 1 | 1 | 0 | 0 | +---+---+---+---+--------+ | 0 | 1 | 1 | 1 | 0 | +---+---+---+---+--------+ | 1 | 0 | 0 | 0 | 0 | +---+---+---+---+--------+ | 1 | 0 | 0 | 1 | 0 | +---+---+---+---+--------+ | 1 | 0 | 1 | 0 | 0 | +---+---+---+---+--------+ | 1 | 0 | 1 | 1 | 0 | +---+---+---+---+--------+ | 1 | 1 | 0 | 0 | 0 | +---+---+---+---+--------+ | 1 | 1 | 0 | 1 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 0 | 0 | +---+---+---+---+--------+ | 1 | 1 | 1 | 1 | 0 | +---+---+---+---+--------+