Expresión (x->y)(y->z)(z->u)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
x⇒y=y∨¬xy⇒z=z∨¬yz⇒u=u∨¬z(x⇒y)∧(y⇒z)∧(z⇒u)=(u∨¬z)∧(y∨¬x)∧(z∨¬y)
(u∨¬z)∧(y∨¬x)∧(z∨¬y)
(u∨(¬z))∧(y∨(¬x))∧(z∨(¬y))
Tabla de verdad
+---+---+---+---+--------+
| u | x | y | z | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 1 |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0 |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 0 |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0 |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 0 |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0 |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 0 |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 0 |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 1 |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 0 |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 1 |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 0 |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0 |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 0 |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 1 |
+---+---+---+---+--------+
Ya está reducido a FNC
(u∨¬z)∧(y∨¬x)∧(z∨¬y)
(u∨(¬z))∧(y∨(¬x))∧(z∨(¬y))
(u∨¬z)∧(y∨¬x)∧(z∨¬y)
(u∨(¬z))∧(y∨(¬x))∧(z∨(¬y))
(u∧y∧z)∨(u∧y∧¬y)∨(u∧z∧¬x)∨(u∧¬x∧¬y)∨(y∧z∧¬z)∨(y∧¬y∧¬z)∨(z∧¬x∧¬z)∨(¬x∧¬y∧¬z)
(u∧y∧z)∨(u∧y∧(¬y))∨(u∧z∧(¬x))∨(y∧z∧(¬z))∨(u∧(¬x)∧(¬y))∨(y∧(¬y)∧(¬z))∨(z∧(¬x)∧(¬z))∨((¬x)∧(¬y)∧(¬z))
(u∧y∧z)∨(u∧¬x∧¬y)∨(¬x∧¬y∧¬z)
(u∧y∧z)∨(u∧(¬x)∧(¬y))∨((¬x)∧(¬y)∧(¬z))