Sr Examen
Lógica matemática/ x=>yz(x->y)(y->=>z)

Expresión x=>yz(x->y)(y->=>z)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

    Ha introducido [src] 
    x⇒(y∧z∧(x⇒y)∧(y⇒z))
    x⇒(y∧z∧(x⇒y)∧(y⇒z))x \Rightarrow \left(y \wedge z \wedge \left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right)\right)
    Solución detallada
    x⇒y=y∨¬xx \Rightarrow y = y \vee \neg x
    y⇒z=z∨¬yy \Rightarrow z = z \vee \neg y
    y∧z∧(x⇒y)∧(y⇒z)=y∧zy \wedge z \wedge \left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right) = y \wedge z
    x⇒(y∧z∧(x⇒y)∧(y⇒z))=(y∧z)∨¬xx \Rightarrow \left(y \wedge z \wedge \left(x \Rightarrow y\right) \wedge \left(y \Rightarrow z\right)\right) = \left(y \wedge z\right) \vee \neg x
    Simplificación [src]
    (y∧z)∨¬x\left(y \wedge z\right) \vee \neg x 
    (¬x)∨(y∧z)
    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 | 0      |
+---+---+---+--------+
| 1 | 0 | 1 | 0      |
+---+---+---+--------+
| 1 | 1 | 0 | 0      |
+---+---+---+--------+
| 1 | 1 | 1 | 1      |
+---+---+---+--------+
    FNC [src]
    (y∨¬x)∧(z∨¬x)\left(y \vee \neg x\right) \wedge \left(z \vee \neg x\right) 
    (y∨(¬x))∧(z∨(¬x))
    FNDP [src]
    (y∧z)∨¬x\left(y \wedge z\right) \vee \neg x 
    (¬x)∨(y∧z)
    FNCD [src]
    (y∨¬x)∧(z∨¬x)\left(y \vee \neg x\right) \wedge \left(z \vee \neg x\right) 
    (y∨(¬x))∧(z∨(¬x))
    FND [src]
    Ya está reducido a FND
    (y∧z)∨¬x\left(y \wedge z\right) \vee \neg x 
    (¬x)∨(y∧z)
    © Sr Examen – Калькулятор