Sr Examen
Lógica matemática/ (a∧b∧c∧x)⇔(b∧x∧(a∨c))

Expresión (a∧b∧c∧x)⇔(b∧x∧(a∨c))

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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