Sr Examen
Expresión ac(¬¬ab+¬c)+¬a(¬abc+¬a)
El profesor se sorprenderá mucho al ver tu solución correcta😉
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Solución
Ha introducido
[src]
((¬a)∧((¬a)∨(b∧c∧(¬a))))∨(a∧c∧(¬((¬c)∨(b∧(¬a)))))
$$\left(\neg a \wedge \left(\left(b \wedge c \wedge \neg a\right) \vee \neg a\right)\right) \vee \left(a \wedge c \wedge \neg \left(\left(b \wedge \neg a\right) \vee \neg c\right)\right)$$
Solución detallada
$$\left(b \wedge c \wedge \neg a\right) \vee \neg a = \neg a$$
$$\neg a \wedge \left(\left(b \wedge c \wedge \neg a\right) \vee \neg a\right) = \neg a$$
$$\neg \left(\left(b \wedge \neg a\right) \vee \neg c\right) = c \wedge \left(a \vee \neg b\right)$$
$$a \wedge c \wedge \neg \left(\left(b \wedge \neg a\right) \vee \neg c\right) = a \wedge c$$
$$\left(\neg a \wedge \left(\left(b \wedge c \wedge \neg a\right) \vee \neg a\right)\right) \vee \left(a \wedge c \wedge \neg \left(\left(b \wedge \neg a\right) \vee \neg c\right)\right) = c \vee \neg a$$
$$\neg a \wedge \left(\left(b \wedge c \wedge \neg a\right) \vee \neg a\right) = \neg a$$
$$\neg \left(\left(b \wedge \neg a\right) \vee \neg c\right) = c \wedge \left(a \vee \neg b\right)$$
$$a \wedge c \wedge \neg \left(\left(b \wedge \neg a\right) \vee \neg c\right) = a \wedge c$$
$$\left(\neg a \wedge \left(\left(b \wedge c \wedge \neg a\right) \vee \neg a\right)\right) \vee \left(a \wedge c \wedge \neg \left(\left(b \wedge \neg a\right) \vee \neg c\right)\right) = c \vee \neg a$$
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 | 0 | +---+---+---+--------+ | 1 | 0 | 1 | 1 | +---+---+---+--------+ | 1 | 1 | 0 | 0 | +---+---+---+--------+ | 1 | 1 | 1 | 1 | +---+---+---+--------+