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