Sr Examen
Lógica matemática/ avbv(c&((av(b')v(c'))'))

Expresión avbv(c&((av(b')v(c'))'))

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    Solución

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