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Lógica matemática/ (¬R˅P→S)&(¬R→¬S)

Expresión (¬R˅P→S)&(¬R→¬S)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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