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Lógica matemática/ C⇔¬(D∨E)

Expresión C⇔¬(D∨E)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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