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Lógica matemática/ aVbVc=(aVb)∧(aVc)

Expresión aVbVc=(aVb)∧(aVc)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    (a∨b∨c)⇔((a∨b)∧(a∨c))
    ((a∨b)∧(a∨c))⇔(a∨b∨c)\left(\left(a \vee b\right) \wedge \left(a \vee c\right)\right) ⇔ \left(a \vee b \vee c\right)
    Solución detallada
    (a∨b)∧(a∨c)=a∨(b∧c)\left(a \vee b\right) \wedge \left(a \vee c\right) = a \vee \left(b \wedge c\right)
    ((a∨b)∧(a∨c))⇔(a∨b∨c)=a∨(b∧c)∨(¬b∧¬c)\left(\left(a \vee b\right) \wedge \left(a \vee c\right)\right) ⇔ \left(a \vee b \vee c\right) = a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right)
    Simplificación [src]
    a∨(b∧c)∨(¬b∧¬c)a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right) 
    a∨(b∧c)∨((¬b)∧(¬c))
    Tabla de verdad 
    +---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1      |
+---+---+---+--------+
| 0 | 0 | 1 | 0      |
+---+---+---+--------+
| 0 | 1 | 0 | 0      |
+---+---+---+--------+
| 0 | 1 | 1 | 1      |
+---+---+---+--------+
| 1 | 0 | 0 | 1      |
+---+---+---+--------+
| 1 | 0 | 1 | 1      |
+---+---+---+--------+
| 1 | 1 | 0 | 1      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FNDP [src]
    a∨(b∧c)∨(¬b∧¬c)a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right) 
    a∨(b∧c)∨((¬b)∧(¬c))
    FNCD [src]
    (a∨b∨¬c)∧(a∨c∨¬b)\left(a \vee b \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right) 
    (a∨b∨(¬c))∧(a∨c∨(¬b))
    FND [src]
    Ya está reducido a FND
    a∨(b∧c)∨(¬b∧¬c)a \vee \left(b \wedge c\right) \vee \left(\neg b \wedge \neg c\right) 
    a∨(b∧c)∨((¬b)∧(¬c))
    FNC [src]
    (a∨b∨¬b)∧(a∨b∨¬c)∧(a∨c∨¬b)∧(a∨c∨¬c)\left(a \vee b \vee \neg b\right) \wedge \left(a \vee b \vee \neg c\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(a \vee c \vee \neg c\right) 
    (a∨b∨(¬b))∧(a∨b∨(¬c))∧(a∨c∨(¬b))∧(a∨c∨(¬c))
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