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