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