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Lógica matemática/ bcdv¬bcdv¬a¬cd

Expresión bcdv¬bcdv¬a¬cd

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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