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Lógica matemática/ неB×(A+C)

Expresión неB×(A+C)

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    Solución

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