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Lógica matemática/ (P↔Q)∨(¬Q→R)

Expresión (P↔Q)∨(¬Q→R)

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    Solución

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