Expresión (XvYv¬(Z&X))&¬(¬(X&Y)v¬Z)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬(x∧z)=¬x∨¬zx∨y∨¬(x∧z)=1¬(x∧y)=¬x∨¬y¬z∨¬(x∧y)=¬x∨¬y∨¬z¬(¬z∨¬(x∧y))=x∧y∧z¬(¬z∨¬(x∧y))∧(x∨y∨¬(x∧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 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 0 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
Ya está reducido a FND
x∧y∧z
x∧y∧z
x∧y∧z
Ya está reducido a FNC
x∧y∧z