Sr Examen
Lógica matemática/ (P∨Q∨R)∧(P∨¬Q∨¬S)

Expresión (P∨Q∨R)∧(P∨¬Q∨¬S)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

    Ha introducido [src] 
    (p∨q∨r)∧(p∨(¬q)∨(¬s))
    $$\left(p \vee q \vee r\right) \wedge \left(p \vee \neg q \vee \neg s\right)$$
    Solución detallada
    $$\left(p \vee q \vee r\right) \wedge \left(p \vee \neg q \vee \neg s\right) = p \vee \left(q \wedge \neg s\right) \vee \left(r \wedge \neg q\right)$$
    Simplificación [src]
    $$p \vee \left(q \wedge \neg s\right) \vee \left(r \wedge \neg q\right)$$ 
    p∨(q∧(¬s))∨(r∧(¬q))
    Tabla de verdad 
    +---+---+---+---+--------+
| p | q | r | s | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 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      |
+---+---+---+---+--------+
    FNC [src]
    $$\left(p \vee q \vee r\right) \wedge \left(p \vee q \vee \neg q\right) \wedge \left(p \vee r \vee \neg s\right) \wedge \left(p \vee \neg q \vee \neg s\right)$$ 
    (p∨q∨r)∧(p∨q∨(¬q))∧(p∨r∨(¬s))∧(p∨(¬q)∨(¬s))
    FNCD [src]
    $$\left(p \vee q \vee r\right) \wedge \left(p \vee \neg q \vee \neg s\right)$$ 
    (p∨q∨r)∧(p∨(¬q)∨(¬s))
    FND [src]
    Ya está reducido a FND
    $$p \vee \left(q \wedge \neg s\right) \vee \left(r \wedge \neg q\right)$$ 
    p∨(q∧(¬s))∨(r∧(¬q))
    FNDP [src]
    $$p \vee \left(q \wedge \neg s\right) \vee \left(r \wedge \neg q\right)$$ 
    p∨(q∧(¬s))∨(r∧(¬q))
    © Sr Examen