Expresión (¬((a&¬b)v(c&¬a)))v(¬(¬(a&¬b)vc)&a)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬(a∧¬b)=b∨¬ac∨¬(a∧¬b)=b∨c∨¬a¬(c∨¬(a∧¬b))=a∧¬b∧¬ca∧¬(c∨¬(a∧¬b))=a∧¬b∧¬c¬((a∧¬b)∨(c∧¬a))=(a∧b)∨(¬a∧¬c)(a∧¬(c∨¬(a∧¬b)))∨¬((a∧¬b)∨(c∧¬a))=(a∧b)∨¬c
(a∧b)∨¬c
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 0 |
+---+---+---+--------+
| 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
(a∧b)∨¬c
(a∧b)∨¬c
(a∨¬c)∧(b∨¬c)
(a∨¬c)∧(b∨¬c)