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Lógica matemática/ (P∧Q)∧¬(R→S)

Expresión (P∧Q)∧¬(R→S)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Solución

Ha introducido [src]

p∧q∧(¬(r⇒s))

$$p \wedge q \wedge r \not\Rightarrow s$$

Solución detallada

$$r \Rightarrow s = s \vee \neg r$$
$$r \not\Rightarrow s = r \wedge \neg s$$
$$p \wedge q \wedge r \not\Rightarrow s = p \wedge q \wedge r \wedge \neg s$$

Simplificación [src]

$$p \wedge q \wedge r \wedge \neg s$$

p∧q∧r∧(¬s)

Tabla de verdad

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

FNC [src]

Ya está reducido a FNC

$$p \wedge q \wedge r \wedge \neg s$$

p∧q∧r∧(¬s)

FNDP [src]

$$p \wedge q \wedge r \wedge \neg s$$

p∧q∧r∧(¬s)

FNCD [src]

$$p \wedge q \wedge r \wedge \neg s$$

p∧q∧r∧(¬s)

FND [src]

Ya está reducido a FND

$$p \wedge q \wedge r \wedge \neg s$$

p∧q∧r∧(¬s)