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Lógica matemática/ pv(rvq)<=>(p^r)v(p^r)

Expresión pv(rvq)<=>(p^r)v(p^r)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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