Sr Examen
Lógica matemática/ [(p∨q)∧(p∨￢q)∧(p∨r)]→[(p∨￢q)∧(￢p∨￢q)∧(￢q∨r)]

Expresión [(p∨q)∧(p∨￢q)∧(p∨r)]→[(p∨￢q)∧(￢p∨￢q)∧(￢q∨r)]

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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