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Lógica matemática/ notbandnotdornotaandbd

Expresión notbandnotdornotaandbd

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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