Expresión not(¬((a→b)and(b→a))and¬c)
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Solución detallada
$$a \Rightarrow b = b \vee \neg a$$
$$b \Rightarrow a = a \vee \neg b$$
$$\left(a \Rightarrow b\right) \wedge \left(b \Rightarrow a\right) = \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$\neg \left(\left(a \Rightarrow b\right) \wedge \left(b \Rightarrow a\right)\right) = \left(a \wedge \neg b\right) \vee \left(b \wedge \neg a\right)$$
$$\neg c \wedge \neg \left(\left(a \Rightarrow b\right) \wedge \left(b \Rightarrow a\right)\right) = \neg c \wedge \left(a \vee b\right) \wedge \left(\neg a \vee \neg b\right)$$
$$\neg \left(\neg c \wedge \neg \left(\left(a \Rightarrow b\right) \wedge \left(b \Rightarrow a\right)\right)\right) = c \vee \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$c \vee \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
Tabla de verdad
+---+---+---+--------+
| a | b | c | result |
+===+===+===+========+
| 0 | 0 | 0 | 1 |
+---+---+---+--------+
| 0 | 0 | 1 | 1 |
+---+---+---+--------+
| 0 | 1 | 0 | 0 |
+---+---+---+--------+
| 0 | 1 | 1 | 1 |
+---+---+---+--------+
| 1 | 0 | 0 | 0 |
+---+---+---+--------+
| 1 | 0 | 1 | 1 |
+---+---+---+--------+
| 1 | 1 | 0 | 1 |
+---+---+---+--------+
| 1 | 1 | 1 | 1 |
+---+---+---+--------+
$$\left(a \vee c \vee \neg b\right) \wedge \left(b \vee c \vee \neg a\right)$$
$$\left(a \vee c \vee \neg a\right) \wedge \left(a \vee c \vee \neg b\right) \wedge \left(b \vee c \vee \neg a\right) \wedge \left(b \vee c \vee \neg b\right)$$
(a∨c∨(¬a))∧(a∨c∨(¬b))∧(b∨c∨(¬a))∧(b∨c∨(¬b))
Ya está reducido a FND
$$c \vee \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$
$$c \vee \left(a \wedge b\right) \vee \left(\neg a \wedge \neg b\right)$$