Expresión ¬(A∧B)∨(C∧B∧A→¬(A∧¬C⇔B∧C))∧¬A∧C∧B∨A
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
¬(a∧b)=¬a∨¬b(a∧¬c)⇔(b∧c)=(c∧¬b)∨(¬a∧¬c)(a∧¬c)≡(b∧c)=(a∧¬c)∨(b∧c)(a∧b∧c)⇒(a∧¬c)≡(b∧c)=1b∧c∧((a∧b∧c)⇒(a∧¬c)≡(b∧c))∧¬a=b∧c∧¬aa∨(b∧c∧((a∧b∧c)⇒(a∧¬c)≡(b∧c))∧¬a)∨¬(a∧b)=1
Tabla de verdad
+---+---+---+--------+
| a | b | c | 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 |
+---+---+---+--------+
