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Lógica matemática/ ((avb)⇒(av¬b))⇔(avc)

Expresión ((avb)⇒(av¬b))⇔(avc)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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