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Lógica matemática/ BC'+B'C'D+BCD'

Expresión BC'+B'C'D+BCD'

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    Solución

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