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Lógica matemática/ bc¬dv¬bc¬dv¬a¬dv¬a¬b¬c¬d

Expresión bc¬dv¬bc¬dv¬a¬dv¬a¬b¬c¬d

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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