Expresión (¬x+¬y&z)⇒((x⇒y)⇒((y+z)⇒¬x))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
x⇒y=y∨¬x(y∨z)⇒¬x=(¬y∧¬z)∨¬x(x⇒y)⇒((y∨z)⇒¬x)=¬x∨¬y((z∧¬y)∨¬x)⇒((x⇒y)⇒((y∨z)⇒¬x))=1
Tabla de verdad
+---+---+---+--------+
| x | y | z | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 1 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+