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Lógica matemática/ (A⇒¬(B∧C⇒¬A)∧C)⇒(A⇒¬B∧C)

Expresión (A⇒¬(B∧C⇒¬A)∧C)⇒(A⇒¬B∧C)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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