Sr Examen
Lógica matemática/ (¬avbv¬c)&(av¬bv¬c)&(¬avbvc)&(¬avbv¬c)&(¬av¬bv¬c)

Expresión (¬avbv¬c)&(av¬bv¬c)&(¬avbvc)&(¬avbv¬c)&(¬av¬bv¬c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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