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Lógica matemática/ [p∨(∼q)]⇔[q∧r]

Expresión [p∨(∼q)]⇔[q∧r]

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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