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Lógica matemática/ BC+A(!C)=(!A)BC+AB(!C)

Expresión BC+A(!C)=(!A)BC+AB(!C)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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