Sr Examen
Lógica matemática/ ¬((¬p)∨¬q)→(r∨¬t)

Expresión ¬((¬p)∨¬q)→(r∨¬t)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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