Expresión xy∨!x(y∨xz)!(x(!y∨z)vyz)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\left(x \wedge \left(z \vee \neg y\right)\right) \vee \left(y \wedge z\right) = \left(x \wedge \neg y\right) \vee \left(y \wedge z\right)$$
$$\neg \left(\left(x \wedge \left(z \vee \neg y\right)\right) \vee \left(y \wedge z\right)\right) = \left(y \wedge \neg z\right) \vee \left(\neg x \wedge \neg y\right)$$
$$\neg x \wedge \neg \left(\left(x \wedge \left(z \vee \neg y\right)\right) \vee \left(y \wedge z\right)\right) \wedge \left(y \vee \left(x \wedge z\right)\right) = y \wedge \neg x \wedge \neg z$$
$$\left(x \wedge y\right) \vee \left(\neg x \wedge \neg \left(\left(x \wedge \left(z \vee \neg y\right)\right) \vee \left(y \wedge z\right)\right) \wedge \left(y \vee \left(x \wedge z\right)\right)\right) = y \wedge \left(x \vee \neg z\right)$$
$$y \wedge \left(x \vee \neg z\right)$$
Tabla de verdad
+---+---+---+--------+
| x | y | z | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
Ya está reducido a FNC
$$y \wedge \left(x \vee \neg z\right)$$
$$y \wedge \left(x \vee \neg z\right)$$
$$\left(x \wedge y\right) \vee \left(y \wedge \neg z\right)$$
$$\left(x \wedge y\right) \vee \left(y \wedge \neg z\right)$$