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Lógica matemática/ ¬((¬P→¬Q)∧(¬Q→R))

Expresión ¬((¬P→¬Q)∧(¬Q→R))

El profesor se sorprenderá mucho al ver tu solución correcta😉

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    Solución

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