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Lógica matemática/ ¬a⇒((p*q)⇒r)

Expresión ¬a⇒((p*q)⇒r)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Solución

Ha introducido [src]

(¬a)⇒((p∧q)⇒r)

$$\neg a \Rightarrow \left(\left(p \wedge q\right) \Rightarrow r\right)$$

Solución detallada

$$\left(p \wedge q\right) \Rightarrow r = r \vee \neg p \vee \neg q$$
$$\neg a \Rightarrow \left(\left(p \wedge q\right) \Rightarrow r\right) = a \vee r \vee \neg p \vee \neg q$$

Simplificación [src]

$$a \vee r \vee \neg p \vee \neg q$$

a∨r∨(¬p)∨(¬q)

Tabla de verdad

+---+---+---+---+--------+
| a | p | q | r | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 1      |
+---+---+---+---+--------+

FND [src]

Ya está reducido a FND

$$a \vee r \vee \neg p \vee \neg q$$

a∨r∨(¬p)∨(¬q)

FNC [src]

Ya está reducido a FNC

$$a \vee r \vee \neg p \vee \neg q$$

a∨r∨(¬p)∨(¬q)

FNDP [src]

$$a \vee r \vee \neg p \vee \neg q$$

a∨r∨(¬p)∨(¬q)

FNCD [src]

$$a \vee r \vee \neg p \vee \neg q$$

a∨r∨(¬p)∨(¬q)