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Lógica matemática/ AD¬C+DCA

Expresión AD¬C+DCA

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    Solución

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