Sr Examen
Lógica matemática/ P→(Q∧R)

Expresión P→(Q∧R)

El profesor se sorprenderá mucho al ver tu solución correcta😉

⌨

    Solución

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