Expresión not(not(x)*y*z+x*not(y)*z)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
(x∧z∧¬y)∨(y∧z∧¬x)=z∧(x∨y)∧(¬x∨¬y)¬((x∧z∧¬y)∨(y∧z∧¬x))=(x∧y)∨(¬x∧¬y)∨¬z
(x∧y)∨(¬x∧¬y)∨¬z
Tabla de verdad
+---+---+---+--------+
| x | y | z | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 0 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
Ya está reducido a FND
(x∧y)∨(¬x∧¬y)∨¬z
(x∨¬x∨¬z)∧(x∨¬y∨¬z)∧(y∨¬x∨¬z)∧(y∨¬y∨¬z)
(x∨(¬x)∨(¬z))∧(x∨(¬y)∨(¬z))∧(y∨(¬x)∨(¬z))∧(y∨(¬y)∨(¬z))
(x∨¬y∨¬z)∧(y∨¬x∨¬z)
(x∨(¬y)∨(¬z))∧(y∨(¬x)∨(¬z))
(x∧y)∨(¬x∧¬y)∨¬z