Sr Examen

Expresión ((¬P∨Q)∧(Q→P))→(P∧¬Q)

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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