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