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