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