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Lógica matemática/ ac∨¬a¬b¬c

Expresión ac∨¬a¬b¬c

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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