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Lógica matemática/ b⊕¬a⇒(¬b+c)⊕c

Expresión b⊕¬a⇒(¬b+c)⊕c

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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