Sr Examen
Lógica matemática/ (s⇒((¬p)∧q)∧r)

Expresión (s⇒((¬p)∧q)∧r)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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