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Lógica matemática/ x=¬(b->¬c)*(¬a+c)

Expresión x=¬(b->¬c)*(¬a+c)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

    Ha introducido [src] 
    x⇔((c∨(¬a))∧(¬(b⇒(¬c))))
    $$x ⇔ \left(b \not\Rightarrow \neg c \wedge \left(c \vee \neg a\right)\right)$$
    Solución detallada
    $$b \Rightarrow \neg c = \neg b \vee \neg c$$
    $$b \not\Rightarrow \neg c = b \wedge c$$
    $$b \not\Rightarrow \neg c \wedge \left(c \vee \neg a\right) = b \wedge c$$
    $$x ⇔ \left(b \not\Rightarrow \neg c \wedge \left(c \vee \neg a\right)\right) = \left(\neg b \wedge \neg x\right) \vee \left(\neg c \wedge \neg x\right) \vee \left(b \wedge c \wedge x\right)$$
    Simplificación [src]
    $$\left(\neg b \wedge \neg x\right) \vee \left(\neg c \wedge \neg x\right) \vee \left(b \wedge c \wedge x\right)$$ 
    (b∧c∧x)∨((¬b)∧(¬x))∨((¬c)∧(¬x))
    Tabla de verdad 
    +---+---+---+---+--------+
| a | b | c | x | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+
    FNDP [src]
    $$\left(\neg b \wedge \neg x\right) \vee \left(\neg c \wedge \neg x\right) \vee \left(b \wedge c \wedge x\right)$$ 
    (b∧c∧x)∨((¬b)∧(¬x))∨((¬c)∧(¬x))
    FNCD [src]
    $$\left(b \vee \neg x\right) \wedge \left(c \vee \neg x\right) \wedge \left(x \vee \neg b \vee \neg c\right)$$ 
    (b∨(¬x))∧(c∨(¬x))∧(x∨(¬b)∨(¬c))
    FND [src]
    Ya está reducido a FND
    $$\left(\neg b \wedge \neg x\right) \vee \left(\neg c \wedge \neg x\right) \vee \left(b \wedge c \wedge x\right)$$ 
    (b∧c∧x)∨((¬b)∧(¬x))∨((¬c)∧(¬x))
    FNC [src]
    $$\left(b \vee \neg x\right) \wedge \left(c \vee \neg x\right) \wedge \left(x \vee \neg x\right) \wedge \left(b \vee \neg b \vee \neg c\right) \wedge \left(b \vee \neg b \vee \neg x\right) \wedge \left(b \vee \neg c \vee \neg x\right) \wedge \left(c \vee \neg b \vee \neg c\right) \wedge \left(c \vee \neg b \vee \neg x\right) \wedge \left(c \vee \neg c \vee \neg x\right) \wedge \left(x \vee \neg b \vee \neg c\right) \wedge \left(x \vee \neg b \vee \neg x\right) \wedge \left(x \vee \neg c \vee \neg x\right)$$ 
    (b∨(¬x))∧(c∨(¬x))∧(x∨(¬x))∧(b∨(¬b)∨(¬c))∧(b∨(¬b)∨(¬x))∧(b∨(¬c)∨(¬x))∧(c∨(¬b)∨(¬c))∧(c∨(¬b)∨(¬x))∧(c∨(¬c)∨(¬x))∧(x∨(¬b)∨(¬c))∧(x∨(¬b)∨(¬x))∧(x∨(¬c)∨(¬x))
    © Sr Examen