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Lógica matemática/ ((AvA)^(BvD)^(BvC))v((BvC)^(CvD))

Expresión ((AvA)^(BvD)^(BvC))v((BvC)^(CvD))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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