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