Expresión ab(¬a⇔a⇒¬a(b⇒a(¬(ab)⇔a)))
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$\neg \left(a \wedge b\right) = \neg a \vee \neg b$$
$$a ⇔ \neg \left(a \wedge b\right) = a \wedge \neg b$$
$$a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right) = a \wedge \neg b$$
$$b \Rightarrow \left(a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right)\right) = \neg b$$
$$\left(b \Rightarrow \left(a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right)\right)\right) \wedge \neg a = \neg a \wedge \neg b$$
$$a \Rightarrow \left(\left(b \Rightarrow \left(a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right)\right)\right) \wedge \neg a\right) = \neg a$$
$$\left(a \Rightarrow \left(\left(b \Rightarrow \left(a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right)\right)\right) \wedge \neg a\right)\right) ⇔ \neg a = 1$$
$$a \wedge b \wedge \left(\left(a \Rightarrow \left(\left(b \Rightarrow \left(a \wedge \left(a ⇔ \neg \left(a \wedge b\right)\right)\right)\right) \wedge \neg a\right)\right) ⇔ \neg a\right) = a \wedge b$$
Tabla de verdad
+---+---+--------+
| a | b | result |
+===+===+========+
| 0 | 0 | 0 |
+---+---+--------+
| 0 | 1 | 0 |
+---+---+--------+
| 1 | 0 | 0 |
+---+---+--------+
| 1 | 1 | 1 |
+---+---+--------+
Ya está reducido a FND
$$a \wedge b$$
Ya está reducido a FNC
$$a \wedge b$$