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Lógica matemática/ ab+¬a¬c

Expresión ab+¬a¬c

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    Solución

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