Sr Examen
Lógica matemática/ (avbvc)&(avbvc)&(¬avbv¬x)

Expresión (avbvc)&(avbvc)&(¬avbv¬x)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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