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Lógica matemática/ (b\cb)ac\(abv(bc\ac))

Expresión (b\cb)ac\(abv(bc\ac))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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