Sr Examen

Expresión (p→q)→r

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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