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Lógica matemática/ ABC⇔(¬A∨BC)⇒(A¬C∨¬BC)

Expresión ABC⇔(¬A∨BC)⇒(A¬C∨¬BC)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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