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Lógica matemática/ а⇒(b∨c)⇔((a⇒b)|(a⇒c))

Expresión а⇒(b∨c)⇔((a⇒b)|(a⇒c))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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