Sr Examen

Expresión (¬(b∧c)→(a∨c))→¬(¬(a∨b)→(c→¬b))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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