Sr Examen

Expresión (∼(p→q)∨~(q→p))∧(∼(q→r)∨~(r→q))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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