Sr Examen
Lógica matemática/ ¬a⇒¬(p∨q)

Expresión ¬a⇒¬(p∨q)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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