Sr Examen

Expresión notz<->not(x->(yvnotz))

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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