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Lógica matemática/ (P∧Q)∧¬(R→S)

Expresión (P∧Q)∧¬(R→S)

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

    Ha introducido [src] 
    p∧q∧(¬(r⇒s))
    $$p \wedge q \wedge r \not\Rightarrow s$$
    Solución detallada
    $$r \Rightarrow s = s \vee \neg r$$
    $$r \not\Rightarrow s = r \wedge \neg s$$
    $$p \wedge q \wedge r \not\Rightarrow s = p \wedge q \wedge r \wedge \neg s$$
    Simplificación [src]
    $$p \wedge q \wedge r \wedge \neg s$$ 
    p∧q∧r∧(¬s)
    Tabla de verdad 
    +---+---+---+---+--------+
| p | q | r | s | result |
+===+===+===+===+========+
| 0 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 0 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 0 | 1 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 0 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 0 | 1 | 0      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 0 | 1      |
+---+---+---+---+--------+
| 1 | 1 | 1 | 1 | 0      |
+---+---+---+---+--------+
    FNC [src]
    Ya está reducido a FNC
    $$p \wedge q \wedge r \wedge \neg s$$ 
    p∧q∧r∧(¬s)
    FNDP [src]
    $$p \wedge q \wedge r \wedge \neg s$$ 
    p∧q∧r∧(¬s)
    FNCD [src]
    $$p \wedge q \wedge r \wedge \neg s$$ 
    p∧q∧r∧(¬s)
    FND [src]
    Ya está reducido a FND
    $$p \wedge q \wedge r \wedge \neg s$$ 
    p∧q∧r∧(¬s)
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