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