Sr Examen
Lógica matemática/ CDB+¬B¬DC

Expresión CDB+¬B¬DC

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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