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Lógica matemática/ PvQvP^Q->!R

Expresión PvQvP^Q->!R

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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