Sr Examen
Lógica matemática/ Bv¬DvA∧D∧¬CvB

Expresión Bv¬DvA∧D∧¬CvB

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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