Expresión CA∨¬(CAB→(¬BA⇔CB))⇔CA∨B¬A
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Solución
Solución detallada
(a∧¬b)⇔(b∧c)=(b∧¬c)∨(¬a∧¬b)(a∧b∧c)⇒((a∧¬b)⇔(b∧c))=¬a∨¬b∨¬c(a∧b∧c)⇒((a∧¬b)⇔(b∧c))=a∧b∧c(a∧c)∨(a∧b∧c)⇒((a∧¬b)⇔(b∧c))=a∧c((a∧c)∨(b∧¬a))⇔((a∧c)∨(a∧b∧c)⇒((a∧¬b)⇔(b∧c)))=a∨¬b
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 0 |
+---+---+---+--------+
| 1 | 0 | 0 | 1 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+