Sr Examen
Lógica matemática/ Bv(-A<->C)

Expresión Bv(-A<->C)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    b∨(¬(a⇔c))
    $$b \vee a \not\equiv c$$
    Solución detallada
    $$a ⇔ c = \left(a \wedge c\right) \vee \left(\neg a \wedge \neg c\right)$$
    $$a \not\equiv c = \left(a \wedge \neg c\right) \vee \left(c \wedge \neg a\right)$$
    $$b \vee a \not\equiv c = b \vee \left(a \wedge \neg c\right) \vee \left(c \wedge \neg a\right)$$
    Simplificación [src]
    $$b \vee \left(a \wedge \neg c\right) \vee \left(c \wedge \neg a\right)$$ 
    b∨(a∧(¬c))∨(c∧(¬a))
    Tabla de verdad 
    +---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 0      |
+---+---+---+--------+
| 0 | 0 | 1 | 1      |
+---+---+---+--------+
| 0 | 1 | 0 | 1      |
+---+---+---+--------+
| 0 | 1 | 1 | 1      |
+---+---+---+--------+
| 1 | 0 | 0 | 1      |
+---+---+---+--------+
| 1 | 0 | 1 | 0      |
+---+---+---+--------+
| 1 | 1 | 0 | 1      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FND [src]
    Ya está reducido a FND
    $$b \vee \left(a \wedge \neg c\right) \vee \left(c \wedge \neg a\right)$$ 
    b∨(a∧(¬c))∨(c∧(¬a))
    FNDP [src]
    $$b \vee \left(a \wedge \neg c\right) \vee \left(c \wedge \neg a\right)$$ 
    b∨(a∧(¬c))∨(c∧(¬a))
    FNCD [src]
    $$\left(a \vee b \vee c\right) \wedge \left(b \vee \neg a \vee \neg c\right)$$ 
    (a∨b∨c)∧(b∨(¬a)∨(¬c))
    FNC [src]
    $$\left(a \vee b \vee c\right) \wedge \left(a \vee b \vee \neg a\right) \wedge \left(b \vee c \vee \neg c\right) \wedge \left(b \vee \neg a \vee \neg c\right)$$ 
    (a∨b∨c)∧(a∨b∨(¬a))∧(b∨c∨(¬c))∧(b∨(¬a)∨(¬c))
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