Sr Examen
Lógica matemática/ notz<->not(x->(notyvz))

Expresión notz<->not(x->(notyvz))

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

    Ha introducido [src] 
    (¬z)⇔(¬(x⇒(z∨(¬y))))
    ¬z⇔x⇏(z∨¬y)\neg z ⇔ x \not\Rightarrow \left(z \vee \neg y\right)
    Solución detallada
    x⇒(z∨¬y)=z∨¬x∨¬yx \Rightarrow \left(z \vee \neg y\right) = z \vee \neg x \vee \neg y
    x⇏(z∨¬y)=x∧y∧¬zx \not\Rightarrow \left(z \vee \neg y\right) = x \wedge y \wedge \neg z
    ¬z⇔x⇏(z∨¬y)=z∨(x∧y)\neg z ⇔ x \not\Rightarrow \left(z \vee \neg y\right) = z \vee \left(x \wedge y\right)
    Simplificación [src]
    z∨(x∧y)z \vee \left(x \wedge y\right) 
    z∨(x∧y)
    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 | 1      |
+---+---+---+--------+
| 1 | 1 | 0 | 1      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FNDP [src]
    z∨(x∧y)z \vee \left(x \wedge y\right) 
    z∨(x∧y)
    FND [src]
    Ya está reducido a FND
    z∨(x∧y)z \vee \left(x \wedge y\right) 
    z∨(x∧y)
    FNCD [src]
    (x∨z)∧(y∨z)\left(x \vee z\right) \wedge \left(y \vee z\right) 
    (x∨z)∧(y∨z)
    FNC [src]
    (x∨z)∧(y∨z)\left(x \vee z\right) \wedge \left(y \vee z\right) 
    (x∨z)∧(y∨z)
    © Sr Examen – Калькулятор