Expresión xyz||xy!z||!xy||x!yz
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(y∧¬x)∨(x∧y∧z)∨(x∧y∧¬z)∨(x∧z∧¬y)=y∨(x∧z)
y∨(x∧z)
Tabla de verdad
+---+---+---+--------+
| x | y | z | result |
+===+===+===+========+
| 0 | 0 | 0 | 0 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
(x∨y)∧(y∨z)
y∨(x∧z)
(x∨y)∧(y∨z)
Ya está reducido a FND
y∨(x∧z)