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Lógica matemática/ ((b⇔¬a)∨¬c)⊕c

Expresión ((b⇔¬a)∨¬c)⊕c

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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