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