Expresión (xvyvz)(-xvyvz)(xv(-y)vz)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(x∨y∨z)∧(x∨z∨¬y)∧(y∨z∨¬x)=z∨(x∧y)
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 |
+---+---+---+--------+
z∨(x∧y)
(x∨z)∧(y∨z)
(x∨z)∧(y∨z)
Ya está reducido a FND
z∨(x∧y)