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Lógica matemática/ AC(-AB+C)+-AC-(A+-BC)

Expresión AC(-AB+C)+-AC-(A+-BC)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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