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Lógica matemática/ BA∨CB∨C~B~A

Expresión BA∨CB∨C~B~A

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    Solución

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