Sr Examen

Expresión zv(y⇒x),x⇒(yv¬z),y&z⇒¬x⇔zv¬x

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    Solución

    Ha introducido [src]
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))

    Вы использовали:
    | - Не-и (штрих Шеффера).
    Возможно вы имели ввиду символ - Дизъюнкция (ИЛИ)?
    Посмотреть с символом ∨
    Solución detallada
    $$y \Rightarrow x = x \vee \neg y$$
    $$z \vee \left(y \Rightarrow x\right) = x \vee z \vee \neg y$$
    $$x \Rightarrow \left(y \vee \neg z\right) = y \vee \neg x \vee \neg z$$
    $$\left(y \wedge z\right) \Rightarrow \neg x = \neg x \vee \neg y \vee \neg z$$
    $$\left(\left(y \wedge z\right) \Rightarrow \neg x\right) ⇔ \left(z \vee \neg x\right) = \left(z \wedge \neg y\right) \vee \neg x$$
    Simplificación [src]
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))
    Tabla de verdad
    +---+---+---+--------+
    | x | y | z | result |
    +===+===+===+========+
    | 0 | 0 | 0 | 1      |
    +---+---+---+--------+
    | 0 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 0 | 1 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 0 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 0 | 1 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 0 | 1      |
    +---+---+---+--------+
    | 1 | 1 | 1 | 1      |
    +---+---+---+--------+
    FNDP [src]
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))
    FND [src]
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))
    FNCD [src]
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))
    FNC [src]
    (Or(z, Implies(y, x)), Implies(x, Or(y, Not(z))), Equivalent(Implies(And(y, z), Not(x)), Or(z, Not(x))))
    (z | (Implies(y, x)), Implies(x, y | ~z), Equivalent(z | ~x, Implies(y & z, ~x)))