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Lógica matemática/ (P↔Q)→R

Expresión (P↔Q)→R

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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