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Lógica matemática/ ¬(¬a∨¬b∧(a∨c)∨b∧((¬(a∨c))))

Expresión ¬(¬a∨¬b∧(a∨c)∨b∧((¬(a∨c))))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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