Sr Examen

Expresión (¬(x↓y)→z)≡y

El profesor se sorprenderá mucho al ver tu solución correcta😉

    Solución

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