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Lógica matemática/ (a∨b)∨(a∧c)=a∨b∧a∧c=a∧b∧(a∨c)=ab∧(a∨c)

Expresión (a∨b)∨(a∧c)=a∨b∧a∧c=a∧b∧(a∨c)=ab∧(a∨c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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